Part A · the equation of motion · spine

Phase portraits: what you know when there is no formula

Almost no equation of motion can be solved in closed form. This unit is the set of questions you can still answer exactly: how many equilibria, of what type, with which basins, and whether anything periodic exists.

needs 01~2 h slides + 4 h problemsproblems: ./mr problems 02

Unit 00 solved one-dimensional motion by quadrature in three cases, and unit 01 proved that the solution exists and is unique. Neither tells you what the solution looks like when no integral can be evaluated, which is the ordinary situation. This unit answers that, and every answer it gives is a theorem rather than a picture.

How to read this deck

One system runs through all six sections: the damped pendulum $\ddot\theta + \gamma\dot\theta + \sin\theta = 0$, which has no closed-form solution for any $\gamma > 0$ and about which we will nevertheless end up knowing almost everything. Words with a dotted underline are glossary terms — click one, or press G for the index. A solid thin underline names a piece of the running example; hover it. Press N for the notes, where the longer proofs are.

Proofs are on the slide when they are short. When they are not, the slide says so and the notes carry them under Full proof. Five boxed results are cited rather than proved — Hartman–Grobman, Poincaré–Bendixson, the Umlaufsatz, the Hopf bifurcation theorem, and Bendixson–Dulac, whose proof waits for the integrals of units 04 and 05. Two more quoted only in the notes, Poincaré–Hopf and Massera's converse, are labelled the same way. Each says where its proof is.

02 · overview

Six questions, one pendulum

the running example

$\ddot\theta + \gamma\dot\theta + \sin\theta = 0$ — a pendulum with linear damping, in units where the only parameter left is $\gamma$. There is no formula for $\theta(t)$. By the end of the deck we will have proved, each by a short route:

Where it can rest, and what a nudge does. Solve $f = 0$: hanging and upright. The trace and determinant of the derivative there say spiral below, saddle above.
It stops hanging down if it starts inside the separatrix. The energy $V$ falls at rate $\gamma\omega^2$, which traps the orbit; $V$ stops falling only where $\omega = 0$, and the one place an orbit can stay with $\omega = 0$ is at rest.
It never repeats itself. Coming back round would return $V$ to its old value, and $V$ only falls.
A torque $T$ breaks it. Rest needs $\sin\theta = T$: at $T = 1$ the two rest points meet and vanish.

Flows. An autonomous equation is a vector field; its solutions partition the state space.
Linearisation. Near an equilibrium, replace $f$ by its derivative — and know exactly when that is a lie.
Lyapunov. Prove stability with a function that decreases, when linearisation cannot decide.
Periodic orbits. Prove one exists without finding it; or prove none does.
The index. An integer attached to a closed curve that counts equilibria.
Bifurcations. What the portrait does as a parameter moves.

Sections 1–3 work in $\mathbb{R}^n$. Sections 4 and 5 are two-dimensional and say so: they are false in three.

The order is not arbitrary. Linearisation is the first tool and the weakest: it answers the local question at a hyperbolic equilibrium and nothing else. Lyapunov functions answer the non-hyperbolic case and, more importantly, give global information — a basin, not a neighbourhood. Poincaré–Bendixson and the index answer the only genuinely global question the plane allows: is there a cycle. Bifurcation theory then puts a parameter on all of it.

Where each route is done. Route 1 is "Linearisation: what it is, and what it claims" (solving $f = 0$) and "The pendulum's two equilibria, classified" (trace and determinant). Route 2 is three slides of section 3: "Differentiating along a solution" computes $\dot V = -\gamma\omega^2$ without solving anything, "LaSalle's invariance principle" turns "nothing can stay where $\dot V = 0$" into a theorem, and "The damped pendulum stops, and we can say where from" applies it on $\{V < 2\}$. Route 3 is the three-line integral of "Ruling cycles out". Route 4 is "The pendulum with a constant torque", where the meeting at $T = 1$ is a saddle-node bifurcation and the index of section 5 explains why the two rest points can only vanish together.

The damping $\gamma$ is dimensionless. Starting from $m l^2 \ddot\theta + b\,\dot\theta + m g l \sin\theta = 0$ and measuring time in units of $\sqrt{l/g}$ (unit 00's time scale), the equation becomes $\ddot\theta + \gamma\dot\theta + \sin\theta = 0$ with $\gamma = b / (m\sqrt{g l^3})$. Check the dimensions: $b$ is a torque per unit angular velocity, $[\mathsf{M}\mathsf{L}^2\mathsf{T}^{-1}]$, and $m\sqrt{gl^3}$ is $\mathsf{M}\sqrt{\mathsf{L}\mathsf{T}^{-2}\mathsf{L}^3} = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-1}$ — the same. For a one-metre pendulum the time unit is $\sqrt{1/9.81} = 0.3193$ s.

section 1

Vector fields, orbits and flows

An autonomous equation is a picture before it is a formula.

02 · flows

No formula, and that is the normal case

Unit 00 reduced $\ddot x = F(x)$ to quadrature by multiplying by $\dot x$. For the pendulum with $\gamma = 0$ that gives

$$\tfrac12\dot\theta^2 - \cos\theta = \text{const},$$

and then an elliptic integral for the period. Put $\gamma > 0$ back in and the trick fails at the first step: $\dot\theta(\ddot\theta + \gamma\dot\theta + \sin\theta)$ is not a total derivative, because $\gamma\dot\theta^2$ is not.

the honest position

This is not a gap in the literature. The set of equations with elementary solutions has measure zero in any reasonable sense, and almost every system in this course after unit 09 is in the complement.

what is still exactly answerable

How many equilibria, and where.
What each looks like nearby — and whether the linear picture is trustworthy.
Which initial conditions end where: basin of attraction.
Whether a periodic solution exists. Not what it is: whether it exists.
How all of that changes with a parameter.

Every one of these is a theorem about $f$ alone, proved without integrating anything. That is the trade this unit offers: less information, all of it certain.

Why $\gamma\dot\theta^2$ blocks the first integral: to integrate $\dot\theta\,\ddot\theta + \gamma\dot\theta^2 + \dot\theta\sin\theta = 0$ term by term you need each term to be $\frac{\mathrm{d}}{\mathrm{d}t}$ of something built from $\theta$ and $\dot\theta$. The first is $\frac{\mathrm{d}}{\mathrm{d}t}(\tfrac12\dot\theta^2)$ and the third is $\frac{\mathrm{d}}{\mathrm{d}t}(-\cos\theta)$, but $\gamma\dot\theta^2$ is not: a function $G(\theta,\dot\theta)$ with $\dot G = \gamma\dot\theta^2$ along every solution would make the energy conserved and the pendulum would never stop, which section 3 will prove it does. Section 3 turns the obstruction into the tool: the quantity that fails to be conserved decreases, and a decreasing quantity is worth more than a conserved one here.

A sharper version of "measure zero": Liouville's theorem on integration in finite terms (Liouville 1833–41; modern account in Rosenlicht, Amer. Math. Monthly 79 (1972) 963) says exactly which elementary functions have elementary antiderivatives, and already $\int e^{-x^2}\mathrm{d}x$ fails. Solving an ODE is strictly harder than integrating. We are not going to be unlucky occasionally; we are going to be unlucky almost always.

02 · flows

An autonomous system is a vector field

definition

An autonomous system on an open $U \subseteq \mathbb{R}^n$ is $\dot{\mathbf{x}} = f(\mathbf{x})$ with $f : U \to \mathbb{R}^n$ locally Lipschitz. Read $f$ as a vector field: at each point of $U$ it names the velocity of whatever passes through.

A solution is a differentiable curve with $\mathbf{x}'(t) = f(\mathbf{x}(t))$; its image is an orbit. The set of all orbits, drawn in $U$, is the phase portrait.

"Autonomous" means $f$ does not depend on $t$ — the law is the same today as tomorrow. Unit 01 turned every $m\ddot{\mathbf{x}} = F(\mathbf{x}, \dot{\mathbf{x}})$ into one of these on $\mathbb{R}^{2n}$, so this is not a restriction on mechanics, it is mechanics.

the running example, as a system

With $\omega = \dot\theta$, the damped pendulum is

$$f(\theta, \omega) = \begin{pmatrix}\omega \\ -\gamma\omega - \sin\theta\end{pmatrix}$$

on $U = \mathbb{R}^2$. It is $C^\infty$, hence locally Lipschitz, so unit 01 applies: one solution through each point.

Nothing here is approximate. The vector field is the equation; a phase portrait is the equation drawn.

Why autonomy matters so much. For a non-autonomous system $\dot{\mathbf x} = f(\mathbf x, t)$ the arrow at a point changes with time, so curves through the same point at different times go different ways and the picture is useless. Autonomy is what makes the phase portrait a well-defined object. Unit 01's trick of appending $\dot t = 1$ turns any non-autonomous system into an autonomous one in one more dimension, at the cost of never having an equilibrium.

Local Lipschitz is enough. A $C^1$ map is locally Lipschitz: on a convex compact subset the mean value inequality gives $|f(x) - f(y)| \le (\sup\|Df\|)|x - y|$. Norton's dome in unit 01 was the reminder that "smooth-looking" is not the same as Lipschitz, and that everything in this unit fails without it: orbits would cross.

Terminology: many books say trajectory for the parametrised solution and orbit for its image. The distinction matters exactly once — a periodic orbit is one set traversed by infinitely many solutions, each a time shift of the others.

02 · flows · example

The pendulum's field, drawn

At each point $(\theta, \omega)$ draw the arrow $f(\theta,\omega) = (\omega,\ -\gamma\omega - \sin\theta)$. That is the whole equation; nothing has been solved.

Three things can be read off with no algebra:

Above the axis everything moves right, below it everything moves left: the first component is $\dot\theta = \omega$.
On the axis the arrows are vertical: there $\omega = 0$, so $\dot\theta = 0$ and only $\omega$ changes.
The arrows vanish at three points, $\theta = 0$ and $\theta = \pm\pi$ with $\omega = 0$ — and $\pm\pi$ are the same state.

A solution is a curve that is tangent to the arrow at every point it passes. The orange one is the only kind of thing "solving" adds.

The damped pendulum's field at $\gamma = \tfrac12$; arrow length grows with speed and is capped. The orange orbit, released near $\theta = -2.45$, follows the arrows into the hanging position.

What to look for in the figure. Follow any arrow with your finger and keep following: that is integrating the equation by eye, and the orange curve is the same thing done numerically. Notice that the arrows on the $\theta$-axis point down for $0 < \theta < \pi$ and up for $-\pi < \theta < 0$ — gravity, $-\sin\theta$, pulling the bob back towards the bottom — and that above and below the axis the damping term $-\gamma\omega$ tilts every arrow towards the axis.

Why the arrows shrink near three points. $|f|$ is the speed in the state space, and it vanishes exactly where $\omega = 0$ and $\sin\theta = 0$. Near those points the motion is slow, and the next sections are about what happens there. The arrow lengths are drawn as an increasing function of $|f|$, capped so that fast regions do not swamp the picture; only the directions are exact.

The vertical arrows on the axis are worth remembering: they are why, on "The damped pendulum stops, and we can say where from", an orbit cannot stay on the line $\omega = 0$ anywhere except at rest.

02 · flows

Orbits cannot cross

theorem

Let $\dot{\mathbf x} = f(\mathbf x)$ be autonomous with $f$ locally Lipschitz. Then

if $\mathbf x(t)$ is a solution, so is $\mathbf x(t + s)$ for every constant $s$;
two orbits are either disjoint or identical.

proof

Put $\mathbf y(t) = \mathbf x(t+s)$. Then $\mathbf y'(t) = \mathbf x'(t+s) = f(\mathbf x(t+s)) = f(\mathbf y(t))$ — this is where autonomy is used, and it is the only place. If two orbits share a point $\mathbf p$, shift both in time so that each passes through $\mathbf p$ at $t = 0$; by uniqueness (unit 01) the two solutions agree on their common interval, and on the union of their maximal intervals by the usual continuation argument. So the orbits coincide. $\square$

what it buys

The orbits partition $U$. In the plane that is a severe constraint: an orbit cannot cross itself or any other, so the picture is forced to be orderly, and section 4 turns that orderliness into an existence theorem for cycles.

two traps

1. Orbits may approach each other without meeting — a spiral into a fixed point never arrives.
2. An equilibrium is an orbit, a one-point one. A curve running into an equilibrium reaches it only as $t \to \infty$, so nothing crosses there either.

The continuation step, in full. Let $\mathbf x_1, \mathbf x_2$ solve the same equation with $\mathbf x_1(0) = \mathbf x_2(0) = \mathbf p$, on maximal intervals $I_1, I_2$. The set $A = \{t \in I_1 \cap I_2 : \mathbf x_1(t) = \mathbf x_2(t)\}$ is closed in $I_1 \cap I_2$ by continuity, non-empty since $0 \in A$, and open: at any $t_0 \in A$, Picard–Lindelöf applied at the point $\mathbf x_1(t_0)$ gives uniqueness on a small interval around $t_0$. $I_1 \cap I_2$ is an interval, hence connected, so $A = I_1 \cap I_2$. Maximality then forces $I_1 = I_2$ — otherwise one extends the other and the shorter was not maximal.

Why "an equilibrium is reached only in infinite time". If $f(\mathbf p) = 0$ then the constant curve is a solution through $\mathbf p$, so by uniqueness it is the only one. A non-constant solution therefore never equals $\mathbf p$ at any finite time. This is exactly what fails on Norton's dome, and the failure is the same one: no Lipschitz condition at the peak.

Note what is not claimed: distinct orbits may be arbitrarily close, and the closure of an orbit may contain other orbits entirely. Section 4's $\omega$-limit sets are made of exactly those.

02 · flows

The flow is a one-parameter group

definition

For $\mathbf x_0 \in U$ let $t \mapsto \varphi_t(\mathbf x_0)$ be the maximal solution with $\varphi_0(\mathbf x_0) = \mathbf x_0$ (unit 01). The family $\{\varphi_t\}$ is the flow of $f$.

theorem · the one-parameter group property

$\varphi_{t+s}(\mathbf x_0) = \varphi_t\big(\varphi_s(\mathbf x_0)\big)$ whenever either side is defined, and $\varphi_0 = \mathrm{id}$. If every solution is defined for all $t$, the map $t \mapsto \varphi_t$ is a group homomorphism $(\mathbb{R}, +) \to \mathrm{Homeo}(U)$, the group under composition of homeomorphisms of $U$ — continuous bijections $U \to U$ with continuous inverse.

proof

Fix $s$ and set $\mathbf u(t) = \varphi_{t+s}(\mathbf x_0)$, $\mathbf v(t) = \varphi_t(\varphi_s(\mathbf x_0))$. By the previous slide $\mathbf u$ solves the equation; so does $\mathbf v$; and $\mathbf u(0) = \mathbf v(0) = \varphi_s(\mathbf x_0)$. Uniqueness gives $\mathbf u = \mathbf v$. Taking $s = -t$ gives $\varphi_{-t} = \varphi_t^{-1}$, so each $\varphi_t$ is a bijection, and continuity of $\varphi_t$ and of its inverse is unit 01's continuous dependence on initial data. $\square$

"whenever defined"

Solutions can blow up in finite time (unit 01), and then $\varphi_t$ is only defined on part of $U$: a local flow, not a group. The pendulum is safe — see the note — and so is every system in this unit that lives on a compact invariant set.

The group property is the reason the phase portrait is a complete description: knowing the picture for one second is knowing it forever, by composition.

It returns in unit 13 as the one-parameter groups Noether's theorem differentiates, and in unit 14 as the Hamiltonian flow, which additionally preserves a volume.

Why the pendulum's solutions are global. $V = \tfrac12\omega^2 + (1-\cos\theta)$ satisfies $\dot V = -\gamma\omega^2 \le 0$, so $V(\mathbf x(t)) \le V(\mathbf x(0))$ for $t \ge 0$, which bounds $|\omega|$ by $\sqrt{2V(0)}$; then $|\theta|$ grows at most linearly. A solution that stays in a compact set for all finite times has maximal interval $[0,\infty)$ — that is unit 01's escape-from-compacta criterion. Backwards in time $V$ rises rather than falls, but not fast enough to escape: $\omega^2 = 2\big(V - (1-\cos\theta)\big) \le 2V$, so $|\mathrm{d}V/\mathrm{d}\tau| = \gamma\omega^2 \le 2\gamma V$ and Gronwall (unit 01) gives $V \le V(0)e^{2\gamma\tau}$ — finite at every finite time, so the orbit is again trapped in a compact set. The flow is complete and $\{\varphi_t\}$ is a genuine group.

Group, precisely. The map $\Phi: \mathbb{R} \to \mathrm{Homeo}(U)$, $t \mapsto \varphi_t$, is a homomorphism into the group of homeomorphisms of $U$ under composition. Its image is an abelian subgroup: $\varphi_t\varphi_s = \varphi_{t+s} = \varphi_s\varphi_t$. It is almost always injective: $\Phi$ has a kernel only if $\varphi_T = \mathrm{id}$ for some $T \ne 0$, which asks every orbit to be periodic with period dividing $T$ — and for the pendulum that happens only in the undamped linear approximation, where every orbit has period $2\pi$.

Continuous dependence gives more than continuity: Gronwall's inequality from unit 01 gives the quantitative bound $|\varphi_t(x) - \varphi_t(y)| \le e^{Lt}|x - y|$ on a set where $f$ is $L$-Lipschitz. That bound grows, and unit 17 is about what happens when the growth is real rather than an artefact of the estimate.

02 · flows · example

The pendulum's state space is a cylinder

Orbits of the damped pendulum at $\gamma = \tfrac12$, integrated numerically for the picture only — every claim on the next slides is proved without them. Dashed red: the separatrix of the undamped system. The two dashed vertical lines are the same line: $\theta$ and $\theta + 2\pi$ are the same state.

The configuration space of a pendulum is a circle, not a line: $\theta$ and $\theta + 2\pi$ are the same physical position with the same velocity, so the state space is $S^1 \times \mathbb{R} = (\mathbb{R}/2\pi\mathbb{Z}) \times \mathbb{R}$, a cylinder. Drawing it as a strip of the plane with the edges identified is a choice of chart, and the strip is what the figure shows.

This matters twice later. In section 5, a loop that goes once round the cylinder cannot be shrunk to a point, so the index theorem — which is about loops bounding a disc — says nothing about it. And a rotating pendulum is a closed orbit on the cylinder that encloses no equilibrium, which on a genuine plane would be impossible. Getting the state space wrong makes both statements come out false.

Reading the figure: every orbit drawn ends at $(0,0)$ or at $(2\pi, 0)$ — the same point on the cylinder, the bob hanging down. The saddles at $\theta = \pm\pi, 3\pi$ are the bob balanced on end. The dashed curve is the $\gamma = 0$ separatrix $V = 2$; with $\gamma = \tfrac12$ the actual orbits cut inwards across it, because $V$ decreases.

The curves were produced by fourth-order Runge–Kutta at step $0.01$ from seven initial conditions. They are illustration, not evidence: a numerical picture can miss a cycle of tiny amplitude or invent one out of step error, which is exactly why the rest of this unit proves things instead.

02 · flows

Two flows with extra structure

a first integral

$E : U \to \mathbb{R}$, non-constant, with $E(\mathbf x(t))$ constant along every solution. Then every orbit lies inside one level set $\{E = c\}$, and in the plane that is the phase portrait: draw the level curves.

a gradient system

$\dot x_i = -\partial V/\partial x_i$. Along a solution $V$ changes at the rate $\dot V = \sum_i (\partial V/\partial x_i)\,\dot x_i$ — the chain rule, proved on "Differentiating along a solution" in section 3 — so $\dot V = -\sum_i (\partial V/\partial x_i)^2 \le 0$, with equality only at equilibria: $V$ decreases along every non-constant orbit, so no periodic orbit and no cycle of any kind can exist.

the pendulum is both, nearly

$\gamma = 0$: $V = \tfrac12\omega^2 + (1-\cos\theta)$ is a first integral, $\dot V = 0$. The orbits are the level curves of $V$, and unit 00's quadrature is the statement that a level curve can be parametrised by an integral.

$\gamma > 0$: $\dot V = -\gamma\omega^2 \le 0$. Not a gradient system — but $V$ still falls, which is all the proofs in section 3 need. Call such a flow gradient-like.

Conserving a function and contracting one are the two structures that make a flow tractable. Unit 14 shows that mechanics without friction always has the first, in a strong form: the flow preserves $H$ and a volume, which is the symplectic structure.

Why a first integral solves a planar system. Level sets of a $C^1$ function $E: \mathbb{R}^2 \to \mathbb{R}$ are curves wherever $DE \ne 0$, and orbits lie in them; since orbits cannot cross, each regular level curve is a single orbit or a chain of them. This is why the undamped pendulum needs no theory: $V = c$ drawn for a range of $c$ is the complete portrait, closed curves inside $\{V < 2\}$ and unbounded ones outside. In three dimensions one first integral leaves a surface, which is still plenty of room — and unit 17 is about what lives there.

Gradient systems have no cycles. Suppose $\mathbf x(t)$ is periodic with period $T$ and not constant. Then $0 = V(\mathbf x(T)) - V(\mathbf x(0)) = \int_0^T \dot V\,\mathrm{d}t = -\int_0^T \sum_i(\partial_i V)^2\,\mathrm{d}t \le 0$, so every $\partial_i V$ vanishes along the orbit and the orbit is a single equilibrium. The identical argument, run in section 4 with $\dot V = -\gamma\omega^2$, kills every periodic orbit of the damped pendulum. Notice it needs no two-dimensionality at all.

A gradient system's orbits cross the level sets of $V$ orthogonally, and its equilibria are the critical points of $V$; Morse theory is the study of what that forces, and it is how the index theorem of section 5 generalises to surfaces.

02 · flows · example

The undamped pendulum, drawn as level curves

With $\gamma = 0$, $V = \tfrac12\omega^2 + (1-\cos\theta)$ is a first integral, so each orbit lies on one curve $V = c$. Solving for $\omega$,

$$\omega = \pm\sqrt{2\,(c - 1 + \cos\theta)} .$$

$0 < c < 2$: a closed curve round the hanging position, defined only where $\cos\theta \ge 1 - c$ — the swing turns back at $\theta = \pm\arccos(1-c)$.
$c = 2$: the separatrix, $\omega = \pm 2\cos(\theta/2)$, running into the upright position.
$c > 2$: $\omega$ never vanishes, so $\theta$ never turns back: the bob goes over the top.

The whole portrait, from one formula and no integration.

Level curves of $V$ for $c = 0.3,\ 0.8,\ 1.4,\ 1.85$ (closed), $2$ (dashed) and $2.6,\ 3.7$ (open). The shaded region $\{V < 2\}$ is where the bob swings; outside it, it rotates.

What to look for. Every curve is traversed clockwise: above the axis $\dot\theta = \omega > 0$ carries the state to the right, below it to the left. The closed curves shrink to the point $(0,0)$ as $c \to 0$ — near the bottom $1 - \cos\theta \approx \theta^2/2$ and the curves are the circles $\theta^2 + \omega^2 = 2c$ of the linear oscillator. As $c \to 2$ they stretch into the eye-shaped separatrix, and the period of the swing grows without bound (unit 00, "The period across every amplitude"), because the separatrix runs into an equilibrium that it reaches only as $t \to \infty$.

How the curves were drawn. For $c < 2$, put $k = \sqrt{c/2}$; then $\theta = 2\arcsin(k\sin\phi)$, $\omega = 2k\cos\phi$ satisfies $V = 2\sin^2(\theta/2) + \tfrac12\omega^2 = 2k^2 = c$ for every $\phi$, which parametrises the closed curve smoothly through its turning points. For $c \ge 2$ the formula on the slide is used directly, with $\cos\theta = 2\cos^2(\theta/2) - 1$. Nothing here is numerical: these are exact level sets.

On the cylinder the open curves close up: $\theta = -\pi$ and $\theta = \pi$ are the same point, so a rotating orbit is a loop round the cylinder. That loop encloses no equilibrium, which matters in section 5.

section 2

Equilibria and linearisation

Replace $f$ by its derivative, and know exactly when that is a lie.

02 · linearisation

The derivative of a map, in one slide

definition

$f : U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $\mathbf a$ if there is a linear map $A : \mathbb{R}^n \to \mathbb{R}^m$ with

$$\begin{gathered}f(\mathbf a + \mathbf h) = f(\mathbf a) + A\mathbf h + r(\mathbf h),\\ |r(\mathbf h)|\,/\,|\mathbf h| \to 0 .\end{gathered}$$

$A$ is unique; write $Df(\mathbf a)$.

theorem · the Jacobian matrix

If $f$ is differentiable at $\mathbf a$ then every partial derivative exists and the matrix of $Df(\mathbf a)$ in the standard bases is $\big(\partial f_i/\partial x_j\big)$.

Proof. Put $\mathbf h = t\mathbf e_j$: $\;\frac{f(\mathbf a + t\mathbf e_j) - f(\mathbf a)}{t} = A\mathbf e_j + \frac{r(t\mathbf e_j)}{t} \to A\mathbf e_j$. The left side is by definition $\partial f/\partial x_j$, so $A\mathbf e_j$ is column $j$. $\square$

the converse needs more

Partial derivatives can all exist at a point where $f$ is not even continuous. Continuous partials do imply differentiability — proved in unit 03, which is where multivariable calculus is done properly. Everything here is $C^\infty$, so the distinction never bites; it is flagged because it is the standard place to go wrong.

for the pendulum

$f(\theta,\omega) = (\omega,\; -\gamma\omega - \sin\theta)$, so

$$Df(\theta,\omega) = \begin{pmatrix} 0 & 1 \\ -\cos\theta & -\gamma\end{pmatrix}.$$

Two entries are constant; the whole $\theta$-dependence is one cosine.

Definition first, matrix second — and this is a habit, not a preference. The derivative is the linear map that best approximates $f$ near $\mathbf a$; the array of partials is what that map looks like once bases are chosen. Everything structural follows from the map (the chain rule is composition of linear maps, as unit 03 shows), and everything computational from the matrix. When unit 16 defines a tensor as a multilinear map and derives its transformation law as a theorem, it is doing the same thing again, and this slide is where the pattern starts.

Uniqueness of $A$. If $A$ and $B$ both work, then $(A-B)\mathbf h = o(|\mathbf h|)$; putting $\mathbf h = t\mathbf v$ and letting $t \to 0$ gives $(A-B)\mathbf v = 0$ for every $\mathbf v$.

The standard counterexample for the warning box: $f(x,y) = xy/(x^2+y^2)$ for $(x,y) \ne 0$ and $f(0,0) = 0$. Both partials at the origin are $0$ (the function vanishes on both axes), yet $f = \tfrac12$ along $y = x$, so $f$ is not continuous at $0$, let alone differentiable.

Dimensions: if $f$ has dimensions $[F]$ and $x_j$ has $[X_j]$, then $\partial f_i/\partial x_j$ has $[F_i]/[X_j]$. For the pendulum every entry of $Df$ has dimensions of inverse time, and in our units time is dimensionless — which is why the matrix can be written with bare numbers at all. Undo the scaling and the eigenvalues carry a factor $\sqrt{g/l}$, as they must.

02 · linearisation

Linearisation: what it is, and what it claims

definition

$\mathbf x^*$ is an equilibrium of $\dot{\mathbf x} = f(\mathbf x)$ if $f(\mathbf x^*) = 0$. Write $\mathbf x = \mathbf x^* + \boldsymbol\xi$; then

$$\dot{\boldsymbol\xi} = f(\mathbf x^* + \boldsymbol\xi) = \underbrace{Df(\mathbf x^*)}_{=:\,A}\boldsymbol\xi + o(|\boldsymbol\xi|).$$

The linearisation at $\mathbf x^*$ is the linear system $\dot{\boldsymbol\xi} = A\boldsymbol\xi$, obtained by deleting the remainder.

deleting the remainder is an assumption

Nothing so far says the two systems behave alike. They do when $A$ is hyperbolic — no eigenvalue with zero real part — and can disagree completely when it is not. Two slides from here, both halves are made precise.

the pendulum's equilibria

$f = 0$ forces $\omega = 0$ and $\sin\theta = 0$: the points $(k\pi, 0)$. On the cylinder there are exactly two: $(0,0)$, hanging down, and $(\pi, 0)$, balanced up. Their Jacobians are

$$\begin{pmatrix}0&1\\-1&-\gamma\end{pmatrix}, \qquad \begin{pmatrix}0&1\\1&-\gamma\end{pmatrix}$$

— the whole difference is the sign of $\cos\theta$, and it decides everything.

Two equations in two unknowns, solved exactly, with no integration. This is the first question of the six and it is always answerable.

Why the equilibria are the right first question. They are the only orbits you can write down for a general $f$, they organise the portrait (everything else flows between them), and they are what bifurcation theory watches. In mechanics an equilibrium of the phase-space system is a configuration of zero velocity and zero force — unit 00 read them off $V'(x) = 0$ in one dimension, and this is the same statement with the phase-space reformulation of unit 01 applied.

The remainder, exactly. $f(\mathbf x^* + \boldsymbol\xi) - A\boldsymbol\xi = r(\boldsymbol\xi)$ with $|r(\boldsymbol\xi)|/|\boldsymbol\xi| \to 0$. For the pendulum at the down position, $-\sin\theta = -\theta + \theta^3/6 - \dots$, so $r$ is $O(|\boldsymbol\xi|^3)$ — cubically small, which is why the linear pendulum is such a good approximation in practice and why unit 00's correction to the period started at $\theta_0^2$.

Counting on the cylinder: on the strip $[-\pi, \pi)$ the equilibria are $\theta = 0$ and $\theta = \pi$ only. Writing "$(k\pi, 0)$, $k \in \mathbb{Z}$" describes the same two points repeated, and a count of equilibria that forgets the identification will disagree with the index theorem in section 5.

02 · linearisation

Planar linear systems: two numbers decide

For $\dot{\boldsymbol\xi} = A\boldsymbol\xi$ with $A$ real $2\times2$, the eigenvalues solve $\lambda^2 - \tau\lambda + \Delta = 0$ with $\tau = \operatorname{tr}A$, $\Delta = \det A$:

$$\lambda_\pm = \tfrac12\Big(\tau \pm \sqrt{\tau^2 - 4\Delta}\Big).$$

So the trace–determinant plane classifies everything. $\Delta < 0$: real eigenvalues of opposite sign, a saddle. $\Delta > 0$ and $\tau^2 > 4\Delta$: a node, stable iff $\tau < 0$. $\tau^2 < 4\Delta$: a spiral, stable iff $\tau < 0$. $\tau = 0 < \Delta$: a centre.

Stability is $\tau < 0$ and $\Delta > 0$ — both conditions, and neither alone.

Where the down equilibrium ($\tau=-\tfrac12$, $\Delta=1$) and the up equilibrium ($\tau=-\tfrac12$, $\Delta=-1$) sit. Moving $\gamma$ slides both horizontally.

Where the quadratic comes from. $\det(A - \lambda I) = \lambda^2 - (\operatorname{tr}A)\lambda + \det A$ for any $2\times2$ matrix: expand. So the eigenvalues are symmetric functions of the two entries that survive a change of basis, and the classification is basis-independent as it must be.

Reading the picture. The parabola $\tau^2 = 4\Delta$ is where the discriminant vanishes: above it the eigenvalues are complex, below it real. The horizontal axis $\Delta = 0$ is where one eigenvalue is zero — $A$ singular, a whole line of equilibria rather than an isolated one. The vertical segment $\tau = 0$, $\Delta > 0$ is the centre case. Those two lines are precisely the non-hyperbolic cases, and they are exactly where the next slide's theorem gives up: hyperbolicity is an open dense condition, and the failures form a set of measure zero that mechanics nevertheless lands on constantly, because conservative systems have $\tau = 0$ identically.

Why $\tau = 0$ for a conservative mechanical system. For $\ddot x = -V'(x)$ the Jacobian is $\begin{pmatrix}0&1\\-V''&0\end{pmatrix}$, trace zero always. Friction is what moves $\tau$ off the axis. So every frictionless equilibrium is either a saddle ($V'' < 0$) or a centre ($V'' > 0$) — and the centre is the case the theory cannot decide, which is the whole reason section 3 exists.

02 · linearisation

Solving $\dot{\boldsymbol\xi} = A\boldsymbol\xi$, all three cases

real, distinct

$A\mathbf v_i = \lambda_i \mathbf v_i$, $\lambda_1 \ne \lambda_2$ real. Then

$\boldsymbol\xi(t) = c_1 e^{\lambda_1 t}\mathbf v_1 + c_2 e^{\lambda_2 t}\mathbf v_2 .$

Differentiate: it solves. The $\mathbf v_i$ are a basis, so any initial condition is met; uniqueness (unit 01) says there is nothing else. Orbits leave along the eigendirection of larger $\lambda$.

repeated

$\lambda$ twice. Either $A = \lambda I$ (a star node, every ray an orbit), or there is $\mathbf w$ with $(A-\lambda I)\mathbf w = \mathbf v$ and

$\boldsymbol\xi(t) = e^{\lambda t}\big(c_1\mathbf v + c_2(\mathbf w + t\mathbf v)\big),$

a degenerate node: one eigendirection, approached by everything.

complex, $\alpha \pm i\beta$

There is a real basis in which $A = \begin{pmatrix}\alpha&-\beta\\\beta&\alpha\end{pmatrix}$ (proof in the notes — no complex exponential needed). In it,

$\boldsymbol\xi(t) = r_0e^{\alpha t}\begin{pmatrix}\cos(\beta t + \phi)\\ \sin(\beta t + \phi)\end{pmatrix}.$

A spiral: $r$ grows by $e^{\alpha t}$, the angle advances at rate $\beta$.

check

Every case is verified, not derived: differentiate the formula, then invoke uniqueness. That is legitimate and it is all we need — the systematic solution of $\dot{\mathbf x} = A\mathbf x$ in $n$ dimensions, by the matrix exponential, is unit 07. Dimensions: $\lambda$, $\alpha$, $\beta$ are all inverse times, so $\lambda t$ is a pure number, as an exponent must be.

Full proof of the real canonical form

Let $A$ be real $2\times2$ with $\tau^2 < 4\Delta$, so it has no real eigenvalue. Set $\alpha = \tau/2$ and $\beta = \sqrt{\Delta - \tau^2/4} > 0$. Pick any $\mathbf p \ne 0$. Then $\{\mathbf p, A\mathbf p\}$ is a basis: if $A\mathbf p = c\mathbf p$ then $c$ would be a real eigenvalue. Every $2\times2$ matrix satisfies its own characteristic polynomial — $A^2 - \tau A + \Delta I = 0$, which for $2\times2$ is a one-line computation — so $A^2\mathbf p = \tau A\mathbf p - \Delta\mathbf p$. Now put $\mathbf q = \beta^{-1}(A\mathbf p - \alpha\mathbf p)$, also a basis vector with $\mathbf p$. Then $A\mathbf p = \alpha\mathbf p + \beta\mathbf q$ by construction, and

$A\mathbf q = \beta^{-1}(A^2\mathbf p - \alpha A\mathbf p) = \beta^{-1}\big((\tau-\alpha)A\mathbf p - \Delta\mathbf p\big) = \beta^{-1}\big(\alpha(\alpha\mathbf p + \beta\mathbf q) - \Delta\mathbf p\big) = \beta^{-1}(\alpha^2-\Delta)\mathbf p + \alpha\mathbf q = -\beta\mathbf p + \alpha\mathbf q,$

using $\alpha^2 - \Delta = -\beta^2$. In the basis $\{\mathbf p, \mathbf q\}$ the matrix of $A$ is therefore $\begin{pmatrix}\alpha&-\beta\\\beta&\alpha\end{pmatrix}$. $\square$ The complex pair $\alpha \pm i\beta$ names the eigenvalues, as linear algebra always has; what does not appear anywhere is the complex exponential $e^{(\alpha+i\beta)t}$, which is unit 06's tool. The real canonical form reaches the same solution without it.

Verifying the spiral solution. With $u = re^{\alpha t}\cos(\beta t + \phi)$, $v = re^{\alpha t}\sin(\beta t+\phi)$: $\dot u = \alpha u - \beta v$ and $\dot v = \beta u + \alpha v$, which is the system. Any initial condition is reached by choosing $r_0$ and $\phi$ in polar form, so uniqueness finishes it.

The repeated case. If $\lambda$ is a double root and $A \ne \lambda I$, then $N = A - \lambda I$ is nonzero with $N^2 = 0$ (Cayley–Hamilton again), so $\operatorname{im}N \subseteq \ker N$ and any $\mathbf w \notin \ker N$ gives $\mathbf v = N\mathbf w \ne 0$ with $N\mathbf v = 0$. Differentiating $e^{\lambda t}(c_1\mathbf v + c_2\mathbf w + c_2 t\mathbf v)$ and using $A\mathbf v = \lambda\mathbf v$, $A\mathbf w = \lambda\mathbf w + \mathbf v$ confirms it solves. This is the $2\times2$ Jordan form, done by hand.

02 · linearisation · example

The cases, drawn on the pendulum's own matrices

Every panel is $\dot{\boldsymbol\xi} = A\boldsymbol\xi$ with $A = \begin{pmatrix}0&1\\\pm1&-\gamma\end{pmatrix}$, the linearisation of the pendulum at the top ($+1$) or the bottom ($-1$). Horizontal: the angle offset $\xi_1$; vertical: the angular velocity $\xi_2$. Dashed: the eigendirections $(1, \lambda)$.

Real distinct roots of opposite sign: a saddle, orbits arrive along one eigenline and leave along the other. Of the same sign: a node, every orbit off the faster eigenline enters along the slower one ($\lambda = -0.38$). A double root with one eigenline: every orbit enters tangent to it. Complex roots: a spiral, the radius shrinking by $0.197$ each turn.

Why these four matrices. They are the cases of the previous slide, and each is a real pendulum: the saddle is the upright position at $\gamma = \tfrac12$; the spiral is the hanging position at the same damping; raising the damping to $\gamma = 2$ (critical damping) merges the two eigenvalues into $-1$, and past it, at $\gamma = 3$, they separate again as two negative reals, $\tfrac12(-3 \pm \sqrt5) = -0.382,\ -2.618$. So the gallery is also a picture of one pendulum's hanging position as the damping is turned up: spiral, degenerate node, node.

Why a node is entered along the slow direction. In $c_1e^{\lambda_1t}\mathbf v_1 + c_2e^{\lambda_2t}\mathbf v_2$ with $\lambda_2 < \lambda_1 < 0$, the second term dies faster, so for large $t$ the solution is dominated by the first and points along $\mathbf v_1$, the eigenvector of the eigenvalue nearer zero. The overdamped pendulum therefore creeps home along the line $\xi_2 = -0.382\,\xi_1$.

The eigenvector $(1, \lambda)$. For $A = \begin{pmatrix}0&1\\c&d\end{pmatrix}$, the first row of $A(1,\lambda)^\top = \lambda(1,\lambda)^\top$ reads $\lambda = \lambda$, and the second, $c + d\lambda = \lambda^2$, is the characteristic equation. So every matrix of this companion shape has eigenvectors $(1, \lambda)$: "angular velocity equals $\lambda$ times angle". The orbits were computed by fourth-order Runge–Kutta from the formulas of the previous slide's system; they agree with those formulas to drawing accuracy.

02 · linearisation · example

The pendulum's two equilibria, classified

down: $\tau = -\gamma$, $\Delta = 1$

$\lambda_\pm = \tfrac12(-\gamma \pm \sqrt{\gamma^2-4})$. For $0 < \gamma < 2$ complex with $\alpha = -\gamma/2 < 0$: a stable spiral. At $\gamma = 2$, a double root $\lambda = -1$: degenerate node, critical damping. For $\gamma > 2$, two negative real roots: a stable node.

At $\gamma = \tfrac12$: $\;\lambda = -0.25 \pm 0.9682\,i$. The angle advances at $\beta = 0.9682$, so a turn takes $2\pi/\beta = 6.489$ time units and the amplitude is multiplied by $e^{-0.25 \times 6.489} = 0.1974$ each turn.

up: $\tau = -\gamma$, $\Delta = -1$

$\Delta < 0$ for every $\gamma$, so always a saddle: one direction leaves, one arrives, and the balanced pendulum is unstable however much damping you add.

At $\gamma = \tfrac12$: $\lambda = 0.7808$ and $-1.2808$. Check: their sum is $-0.5 = \tau$ and their product is $-1.0000 = \Delta$.

back to seconds

For $l = 1$ m, one time unit is $\sqrt{l/g} = 0.3193$ s. The small undamped period $2\pi$ is $2.006$ s; at $\gamma=\tfrac12$ the damped period $6.489$ is $2.072$ s — damping makes it slower, by $3.3\%$.

Every number here. $\gamma = \tfrac12$: $\gamma^2 - 4 = -3.75$, so $\lambda = -0.25 \pm \tfrac12\sqrt{3.75}\,i = -0.25 \pm 0.96825\,i$, and $\beta = 0.96825$, $2\pi/\beta = 6.48925$, $e^{-0.25\cdot 6.48925} = 0.19744$. At the up position, $\lambda = \tfrac12(-0.5 \pm \sqrt{0.25+4}) = \tfrac12(-0.5 \pm 2.06155)$, giving $0.78078$ and $-1.28078$; their product is $-1.00000$ and their sum $-0.5$. Seconds: $\sqrt{1/9.81} = 0.31928$ s; $2\pi \times 0.31928 = 2.00607$ s; $6.48925 \times 0.31928 = 2.07186$ s; the ratio is $1.0328$.

The saddle's eigendirections. For $A = \begin{pmatrix}0&1\\1&-\gamma\end{pmatrix}$ an eigenvector for $\lambda$ is $(1, \lambda)$, since the first row reads $0\cdot1 + 1\cdot\lambda = \lambda\cdot1$. So the unstable direction at the top is $(1, 0.7808)$ and the stable one $(1, -1.2808)$: to balance the pendulum you must give it a velocity towards the upright position, in exactly that ratio. Anything else and the saddle's unstable direction takes over. The stable direction is the separatrix of the damped system, and it is a one-dimensional set of initial conditions in a two-dimensional space — hence unobservable in practice, which is what "unstable" means physically.

Critical damping at $\gamma = 2$ is $b = 2m\sqrt{gl^3}$, the classic door-closer condition: return without overshoot, in the shortest time. It is the boundary between spiral and node, and unit 06 does the driven version of this whole calculation properly.

02 · linearisation

Hartman–Grobman, and the hole in it

theorem (Hartman 1960) · cited

Let $f$ be $C^1$ near an equilibrium $\mathbf x^*$ and let $A = Df(\mathbf x^*)$ be hyperbolic: $\operatorname{Re}\lambda \ne 0$ for every eigenvalue. Then there are neighbourhoods $U \ni \mathbf x^*$, $W \ni 0$ and a homeomorphism $h : U \to W$ carrying orbits of $\dot{\mathbf x} = f(\mathbf x)$ onto orbits of $\dot{\boldsymbol\xi} = A\boldsymbol\xi$, preserving the time parameter — a topological conjugacy.

not proved here: Perko, Differential Equations and Dynamical Systems, §2.8

So at a hyperbolic equilibrium the linear picture is the true picture, up to a continuous change of coordinates: saddles stay saddles, sinks stay sinks, and nothing new appears nearby.

what it does not say

Centres are excluded. $\operatorname{Re}\lambda = 0$ is precisely the case left out, and it is not an oversight: the next slide shows two systems with the same linearisation $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, one with every orbit spiralling in, one with every orbit spiralling out.

$h$ is only a homeomorphism. It need not be differentiable, so it does not preserve eigenvalues or rates — only the qualitative picture. Making $h$ smooth needs non-resonance conditions (Sternberg's theorem).

Every frictionless mechanical equilibrium that is not a saddle is a centre. The exclusion is not an edge case in this course; it is half of it.

Why hyperbolicity is the right hypothesis. If $\operatorname{Re}\lambda \ne 0$ for all eigenvalues, the linear flow expands in some directions and contracts in others at definite exponential rates, and the $o(|\boldsymbol\xi|)$ remainder is too small to compete near the equilibrium. If some $\operatorname{Re}\lambda = 0$ there is a direction along which the linear flow neither grows nor decays, and then the remainder — however small — decides the behaviour by itself. That is not a defect of the proof; it is the truth, as the next slide demonstrates with an explicit family.

What "preserving the time parameter" means. Writing $\psi_t$ for the flow of the linear system — the one written down by hand on "Solving $\dot{\boldsymbol\xi} = A\boldsymbol\xi$, all three cases", without any matrix exponential — the requirement is $h(\varphi_t(\mathbf x)) = \psi_t(h(\mathbf x))$ for all small $t$ and $\mathbf x \in U$. Without it one could conjugate a sink to a source by reversing time, and the theorem would be worthless.

History. Proved by Hartman in 1960 and independently by Grobman in 1959. The proof is a fixed-point argument in a space of bounded continuous maps — the same Banach fixed-point theorem unit 01 used for Picard–Lindelöf, applied to a different operator. It is not long, but it needs the stable manifold theorem's machinery to be set up first, and this course does not need the machinery elsewhere, so it is cited rather than proved. That is the only kind of citation this course allows: a named theorem with a stated proof location.

02 · linearisation · example

Hartman–Grobman at the pendulum's saddle

The upright equilibrium at $\gamma = \tfrac12$ is hyperbolic ($\lambda = 0.7808,\ -1.2808$), so the theorem applies. Left, the true orbits near it; right, the orbits of its linearisation. Orange: the orbits that run straight into the equilibrium or straight out of it.

Same pattern, different shape. The conjugacy $h$ straightens the curved orange orbits onto the eigenlines and carries every other orbit with them, timing included. Near the centre the two pictures agree to first order; further out the curvature is the $o(|\boldsymbol\xi|)$ remainder at work.

What to look for. In both panels four orbits meet the equilibrium: two arriving (along the stable direction $(1, -1.2808)$ at the centre) and two leaving (along $(1, 0.7808)$). On the left these four are curved, because the pendulum's restoring term is $-\sin\theta$, not its linear part; on the right they are the straight eigenlines. Every other orbit swerves past the saddle in the same way in both panels. That shared pattern — and nothing about the curvature — is what Hartman–Grobman promises.

How the left panel was drawn. In the offset $u = \theta - \pi$, the pendulum is $\dot u = \omega$, $\dot\omega = -\gamma\omega + \sin u$, since $-\sin(\pi + u) = \sin u$. The orange curves were found by starting a distance $10^{-5}$ from the equilibrium along each eigenvector and integrating forward (the outgoing pair) or backward in time (the incoming pair); the incoming pair is the separatrix of the damped pendulum, which "The pendulum's two equilibria, classified" described in its notes. The other orbits start near the incoming curves and are integrated forward by fourth-order Runge–Kutta.

What $h$ is not asked to do. The figure makes $h$ look like a gentle bending, but the theorem promises only continuity, and in general no more is true: a differentiable conjugacy would match the rates of approach along corresponding orbits, and a merely continuous one need not. That is the second warning on "Hartman–Grobman, and the hole in it".

02 · linearisation · example

When linearisation lies

one parameter, three different truths

For $a \in \mathbb{R}$ consider

$$\dot x = -y + a\,x(x^2+y^2), \qquad \dot y = x + a\,y(x^2+y^2).$$

The Jacobian at the origin is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ for every $a$: $\tau = 0$, $\Delta = 1$, a centre. Now put $r^2 = x^2+y^2$ and differentiate:

$$r\dot r = x\dot x + y\dot y = a(x^2+y^2)^2 = a r^4 \;\Longrightarrow\; \dot r = a r^3 .$$

So $r$ grows for $a > 0$, shrinks for $a < 0$, and is constant for $a = 0$. The origin is an unstable spiral, an asymptotically stable spiral, and a centre — with identical linearisations.

check the algebra

$x\dot x + y\dot y = x(-y + axr^2) + y(x + ayr^2) = -xy + xy + a r^2(x^2+y^2) = ar^4$. The linear parts cancel exactly — that is what $\tau = 0$ means — and the cubic terms are what is left.

The next slide draws all three.

Solving it completely, for once. The angle satisfies $\dot\theta = (x\dot y - y\dot x)/r^2 = (x^2+y^2)/r^2 = 1$ — the cubic terms are radial and contribute nothing to the rotation. So $\theta = t + \theta_0$ and $\dot r = ar^3$ separates (unit 00): $r(t) = r_0(1 - 2ar_0^2 t)^{-1/2}$. For $a > 0$ this blows up at $t = 1/(2ar_0^2)$, a finite time — unit 01's blow-up, met again. For $a < 0$ it decays like $t^{-1/2}$, which is not exponential: the approach to the origin is slower than any hyperbolic sink, another sign that the linearisation is not in charge.

Dimensional sanity. With $a < 0$, $|a|r_0^2$ has dimensions of inverse time, so $t^{-1/2}$ decay carries the right scale. The $1$ in $\dot\theta = 1$ is the linear frequency, and the whole family has the same one.

02 · linearisation · example

Three systems, one linearisation

$\dot x = -y + a\,x r^2$, $\dot y = x + a\,y r^2$ with $a = -1,\ 0,\ 1$, from the exact solution $r = r_0(1 - 2ar_0^2t)^{-1/2}$. Left: from $r_0 = 1.1$, 40 time units only bring it to $r = 0.11$ — decay like $t^{-1/2}$. Right: from $r_0 = 0.15$ it leaves the frame at $t = 21.8$ and reaches infinity at $t = 22.2$.

the lesson, precisely

All three linearise to the middle panel. A centre in the linearisation is no information at all about the nonlinear system; section 3 supplies the instrument that decides — for $a < 0$, $V = x^2+y^2$ decreases, and that settles it. Problem P4 is the same trap with $\dot x = -y-x^3$, $\dot y = x-y^3$.

What to look for. The three panels differ only in the cubic term, which is invisible to the derivative at the origin. On the left the spiral tightens slowly: each turn takes the same time $2\pi$, and the radius after $n$ turns is $1.1/\sqrt{1 + 2.42 \cdot 2\pi n}$ — the turns crowd together instead of shrinking by a fixed factor, as a hyperbolic spiral's would (compare the spiral panel of "The cases, drawn on the pendulum's own matrices", which loses a factor $0.197$ per turn). On the right the spiral opens faster and faster and the solution blows up in finite time.

The numbers in the caption. For $a = -1$, $r_0 = 1.1$, $t = 40$: $r = 1.1/\sqrt{1 + 2\cdot1.21\cdot40} = 1.1/\sqrt{97.8} = 0.111$. For $a = 1$, $r_0 = 0.15$: blow-up at $t = 1/(2 \cdot 0.0225) = 22.2$; the frame edge $r = 1.14$ is reached at $t = (1 - (0.15/1.14)^2)/0.045 = 21.8$.

This example is the reason the course insists on stating Hartman–Grobman with its hypothesis rather than as "linearise and look". A student who has only ever met hyperbolic examples will believe the linearisation always; a student who has met this one will not. Note also that the three cases here are not exotic: $a$ passing through zero is a Hopf bifurcation, section 6's last topic, and the family is its normal form.

section 3

Lyapunov's direct method

Prove stability with a function that falls, and get a basin for free.

02 · lyapunov

Stability, defined

definition (Lyapunov, 1892)

Let $\mathbf x^*$ be an equilibrium. It is

stable if for every $\varepsilon > 0$ there is $\delta > 0$ such that $|\mathbf x_0 - \mathbf x^*| < \delta$ implies $|\varphi_t(\mathbf x_0) - \mathbf x^*| < \varepsilon$ for all $t \ge 0$;
asymptotically stable if stable and, in addition, $\varphi_t(\mathbf x_0) \to \mathbf x^*$ for all $\mathbf x_0$ near enough;
unstable otherwise.

both halves are needed

Attraction does not imply stability: there are systems in which every orbit converges to $\mathbf x^*$ but some first travel a long way away. Stability is the uniform statement, and it is the one that matters physically.

Stability: whatever $\varepsilon$ you are given, some $\delta$ works. Asymptotic stability additionally sends the orbit to the centre.

The order of quantifiers is the definition. $\forall\varepsilon\,\exists\delta\,\forall \mathbf x_0\,\forall t \ge 0$. Swapping the first two gives a false statement (one $\delta$ for all $\varepsilon$), and dropping "$\forall t$" gives nothing at all. Write it out in full once; every stability proof in this course is an instance of it.

An attracting but unstable equilibrium. In polar coordinates take $\dot r = r(1-r)$, $\dot\theta = \sin^2(\theta/2)$. On the circle $r = 1$ the angular equation has an equilibrium at $\theta = 0$ that every other point approaches by going all the way round. Adding the radial equation, every orbit except the one sitting at the origin converges to $(r,\theta)=(1,0)$, yet a point just past $\theta = 0$ must travel the full circle to return: not stable. Vinograd's example is the classical version of this. (This is a statement about an equilibrium of the whole plane after the standard polar-to-Cartesian check; the point of citing it is only that the two halves of the definition are independent.)

Why the flow, not the formula. The definition mentions $\varphi_t$ and nothing else, so it can be checked for systems nobody can solve — which is the whole programme of this unit. Lyapunov's 1892 thesis, The General Problem of the Stability of Motion, introduced both the definition and the method on the next slides; it was written in Russian, translated into French in 1907, and largely ignored in the West until the 1940s.

02 · lyapunov

Differentiating along a solution

theorem · the orbital derivative

Let $V : U \to \mathbb{R}$ be differentiable and $\mathbf x(t)$ a solution of $\dot{\mathbf x} = f(\mathbf x)$. Then $t \mapsto V(\mathbf x(t))$ is differentiable, with derivative $\dot V(\mathbf x(t))$, where

$$\dot V(\mathbf x) \;:=\; DV(\mathbf x)\,f(\mathbf x) = \sum_{i=1}^n \partial_i V(\mathbf x)\,f_i(\mathbf x).$$

The right-hand side is a function of position alone: no solving required.

proof

Write $\mathbf x(t+s) - \mathbf x(t) = s\,f(\mathbf x(t)) + \boldsymbol\rho(s)$ with $\boldsymbol\rho(s)/s \to 0$. Differentiability of $V$ at $\mathbf x(t)$ gives $V(\mathbf x(t+s)) - V(\mathbf x(t)) = DV(\mathbf x(t))\big(s f + \boldsymbol\rho(s)\big) + r$, where $|r|/|s f + \boldsymbol\rho(s)| \to 0$ and $|sf + \boldsymbol\rho(s)| = O(s)$. Divide by $s$ and let $s \to 0$: the $\boldsymbol\rho$ and $r$ terms vanish. $\square$

on the pendulum

With $V = \tfrac12\omega^2 + (1-\cos\theta)$ and $f = (\omega,\,-\gamma\omega-\sin\theta)$:

$$\begin{aligned}\dot V &= \underbrace{\sin\theta}_{\partial_\theta V}\,\omega + \underbrace{\omega}_{\partial_\omega V}\,(-\gamma\omega - \sin\theta)\\ &= -\gamma\omega^2 .\end{aligned}$$

Three lines, no solution of anything, and it is exact: the energy of the damped pendulum falls at rate $\gamma\omega^2$, always.

checks

Dimensions: $V$ is an energy over $mgl$, $\dot V$ an inverse time; $\gamma\omega^2$ is $\text{(number)}\times\text{(inverse time)}^2$ — and one power of inverse time is the time unit itself, so both sides match. Limit: $\gamma\to0$ gives $\dot V = 0$, the conservation of unit 00.

What the theorem really is. It is the multivariable chain rule for the special case of a curve composed with a scalar function; unit 03 proves the general chain rule for $D(g\circ f) = Dg\,Df$ as composition of linear maps, of which this is the case $\mathbb{R} \to \mathbb{R}^n \to \mathbb{R}$. The proof above is written out because this unit needs it now and cannot wait a unit for it.

Why $\dot V$ is computable without solving. The formula $\dot V(\mathbf x) = \sum_i \partial_i V(\mathbf x) f_i(\mathbf x)$ contains no $t$. You evaluate it at a point and learn whether $V$ is rising or falling there, for every orbit through it at once. That is the entire trick of the direct method, and the word "direct" in Lyapunov's name for it means exactly "without integrating the equations".

On notation. $DV$ is a linear map $\mathbb{R}^n \to \mathbb{R}$ — a row, a covector — and $\dot V = DV \cdot f$ pairs it with a vector. The gradient $\nabla V$, which unit 04 introduces, is the vector you get from $DV$ after choosing an inner product; they have the same components in an orthonormal basis and different ones otherwise. The distinction is invisible here and unavoidable from unit 28 onward, so the pairing is written as a row acting on a column deliberately.

02 · lyapunov

Lyapunov's theorem

theorem

Let $\mathbf x^*$ be an equilibrium and $N$ a neighbourhood of it. Suppose $V \in C^1(N)$ satisfies

$V(\mathbf x^*) = 0$ and $V > 0$ on $N \setminus \{\mathbf x^*\}$ (positive definite);
$\dot V \le 0$ on $N$.

Then $\mathbf x^*$ is stable. If moreover $\dot V < 0$ on $N \setminus \{\mathbf x^*\}$, it is asymptotically stable. Such a $V$ is a Lyapunov function.

proof of stability

Given $\varepsilon$ with $\overline B_\varepsilon(\mathbf x^*) \subset N$, let $m = \min\{V(\mathbf x) : |\mathbf x - \mathbf x^*| = \varepsilon\}$, which is $>0$ by compactness of the sphere and positivity of $V$. By continuity pick $\delta \in (0,\varepsilon)$ with $V < m$ on $B_\delta$. For $\mathbf x_0 \in B_\delta$, $V(\varphi_t(\mathbf x_0)) \le V(\mathbf x_0) < m$ as long as the orbit stays in $B_\varepsilon$ — so it can never reach the sphere, where $V \ge m$. $\square$

asymptotic part in the notes (N)

why this is the better instrument

It needs no eigenvalues, no hyperbolicity and no linear algebra. It works at the centres Hartman–Grobman excludes. And its conclusion is not local: if $\{V < c\}$ is bounded and $\dot V \le 0$ on it, the whole sublevel set is an invariant region — a basin estimate, which linearisation can never supply.

the catch

There is no algorithm for finding $V$. In mechanics there is a standard first guess — the energy — and it works remarkably often, because dissipation is exactly the statement that energy falls.

Full proof of the asymptotic part

Assume $\dot V < 0$ on $N\setminus\{\mathbf x^*\}$ and take $\mathbf x_0 \in B_\delta$ as in the stability proof, so the orbit stays in $\overline B_\varepsilon$ for all $t \ge 0$ and exists for all $t \ge 0$ (unit 01: a solution remaining in a compact subset of $U$ has maximal interval $[0,\infty)$). Then $t \mapsto V(\varphi_t(\mathbf x_0))$ is non-increasing and bounded below by $0$, so it converges to some $c \ge 0$. Suppose $c > 0$. The set $K = \{\mathbf x \in \overline B_\varepsilon : c \le V(\mathbf x) \le V(\mathbf x_0)\}$ is compact and does not contain $\mathbf x^*$ (where $V = 0 < c$), so $-k := \max_K \dot V < 0$ by continuity of $\dot V$ and compactness. The orbit stays in $K$ for all $t \ge 0$, hence $V(\varphi_t(\mathbf x_0)) \le V(\mathbf x_0) - kt \to -\infty$, contradicting $V \ge 0$. So $c = 0$.

Finally $V(\varphi_t) \to 0$ forces $\varphi_t \to \mathbf x^*$: if not, there are $\eta > 0$ and $t_n \to \infty$ with $|\varphi_{t_n} - \mathbf x^*| \ge \eta$; those points lie in the compact set $\overline B_\varepsilon \setminus B_\eta$, so a subsequence converges to some $\mathbf y$ there, and continuity gives $V(\mathbf y) = 0$ with $\mathbf y \ne \mathbf x^*$ — contradicting positive definiteness. $\square$

Where each hypothesis is used. Positive definiteness makes the sublevel sets shrink around $\mathbf x^*$ (used twice, and it is what fails if $V$ has a saddle at $\mathbf x^*$). $\dot V \le 0$ makes sublevel sets forward invariant. Compactness of spheres in $\mathbb{R}^n$ is used for $m > 0$ and for $-k < 0$; in infinite dimensions both steps fail, which is why the infinite-dimensional theory is a different subject.

A converse holds — an asymptotically stable equilibrium always admits a Lyapunov function — so nothing is lost by the method, only by our ingenuity in guessing. Cited: Massera, "On Liapounoff's conditions of stability", Ann. of Math. 50 (1949) 705–721.

02 · lyapunov · example

Lyapunov's proof, as a picture

The stability proof in three moves, drawn for the pendulum's energy $V$ with $\gamma = \tfrac12$:

1. On the circle $|\mathbf x| = \varepsilon$ (here $\varepsilon = 2$), $V$ has a minimum $m > 0$: here $m = 1 - \cos 2 = 1.416$, at $(\pm 2, 0)$.

2. The region $\{V < m\}$ (shaded) contains a ball $|\mathbf x| < \delta$; here $\delta = \sqrt{2m} = 1.683$.

3. An orbit from $\mathbf x_0$ in that ball can only go down in $V$, so it never reaches the orange curve $V = m$ — and the circle lies outside that curve.

The orbit crosses each level curve of $V$ inwards and never outwards, because $\dot V \le 0$. It is trapped inside the shaded region, hence inside the $\varepsilon$-circle.

Every number in the picture. On $|\mathbf x| = 2$ write $(\theta, \omega) = 2(\cos\phi, \sin\phi)$; then $V = 2\sin^2\phi + 1 - \cos(2\cos\phi)$, minimised over $\phi$ numerically at $\phi = 0$, where $V = 1 - \cos 2 = 1.4161$. For $\delta$: since $1 - \cos\theta \le \theta^2/2$, $V \le \tfrac12(\theta^2 + \omega^2)$, so $V < m$ whenever $|\mathbf x| < \sqrt{2m} = 1.6829$; and on the $\omega$-axis $V = \tfrac12\omega^2$ reaches $m$ exactly at $|\omega| = \sqrt{2m}$, so no larger $\delta$ works. The orbit starts at $(1.05, 1.05)$, where $|\mathbf x| = 1.485 < \delta$ and $V = 1.054 < m$.

What the picture shows that the proof does not need. The orbit actually spirals all the way in — the damped pendulum's bottom is asymptotically stable — but the stability proof uses only that it stays inside. Getting convergence from this $V$ needs more than Lyapunov's theorem, because $\dot V = -\gamma\omega^2$ vanishes on the whole axis $\omega = 0$; that is the job of "LaSalle's invariance principle", later in this section.

Why the level curves are not circles. $V$ is not a norm; its level curves are wider in $\theta$ than in $\omega$ because $1 - \cos\theta$ grows more slowly than $\theta^2/2$. That is exactly why the proof has to pass through $m$ and $\delta$ rather than use one radius throughout: balls and sublevel sets are different shapes, and the argument nests one inside the other twice.

02 · lyapunov · example

The undamped pendulum: a centre, proved stable

$\gamma = 0$

Linearisation at $(0,0)$: $\tau = 0$, $\Delta = 1$ — a centre, and by the two slides above that is no information. Take $V = \tfrac12\omega^2 + (1-\cos\theta)$ on $N = \{|\theta| < \pi\} \times \mathbb{R}$:

$V(0,0) = 0$, and $V > 0$ elsewhere in $N$ since $1 - \cos\theta > 0$ for $0 < |\theta| < \pi$;
$\dot V = -\gamma\omega^2 = 0 \le 0$.

Lyapunov's theorem applies: the hanging position is stable. Not asymptotically — $\dot V$ is not negative anywhere — and indeed it is not, since the orbits are the closed curves $V = c$.

this is a proof, and it is sharp

It gives exactly what is true and nothing more. The same $V$ with the same argument fails to show asymptotic stability, and asymptotic stability is false. A method that had "proved" it would be wrong.

and it settles the previous slide's family

For $\dot x = -y + a x r^2$, $\dot y = x + a y r^2$ with $a < 0$, take $V = x^2+y^2$. Then $\dot V = 2ar^4 < 0$ off the origin, so the origin is asymptotically stable. With $a = 0$, $\dot V = 0$: stable, not asymptotically. With $a > 0$, $\dot V > 0$: unstable. One function, all three cases, where the linearisation could not distinguish them.

Why $N$ must exclude $\theta = \pm\pi$. $V(\pm\pi, 0) = 2 \ne 0$, so positivity is not the problem — but if $N$ contained the whole cylinder, the sublevel set $\{V < c\}$ for $c > 2$ would wrap all the way round and no longer shrink to the origin, and the $\delta$ produced by the proof would be useless. Keeping $|\theta| < \pi$ keeps the sublevel sets compact and centred, which is exactly what the proof consumes.

A physicist's reading. Total energy is conserved and has a strict minimum at the hanging position; therefore a state of small energy stays a state of small energy, and small energy means small displacement and small speed. That is Dirichlet's stability theorem (1846), which predates Lyapunov and is the special case $V = $ energy, $\dot V = 0$, of the theorem on the previous slide. Lyapunov's advance was to see that the function need not be the energy and need not be conserved.

Why $\dot V > 0$ gives instability, since the theorem opposite does not say so. Fix $\varepsilon$ and start at $\mathbf x_0 \ne 0$ with $a := V(\mathbf x_0) > 0$ inside $B_\varepsilon$. While the orbit stays in $B_\varepsilon$ it stays in the compact set $\{a \le V \le \varepsilon^2\}$, which misses the origin, so $\dot V \ge k > 0$ there by continuity and compactness; then $V \ge a + kt$ passes $\varepsilon^2$ in finite time and the orbit must have left $B_\varepsilon$. This works for every $a$, however small, which is exactly instability. Problem P4 runs the same argument on a harder pair.

Sharpness, once more. A centre in the linearisation is compatible with stable, asymptotically stable and unstable. The energy argument here decides which, in the one case where mechanics hands us the function. When it does not — the damped case, where the energy is not conserved — LaSalle two slides on rescues the argument.

02 · lyapunov

Where orbits end up: $\omega$-limit sets

definition

For a forward orbit define its omega-limit set

$$\begin{aligned}\omega(\mathbf x_0) = \{\mathbf y :\ &\varphi_{t_n}(\mathbf x_0) \to \mathbf y\\ &\text{for some } t_n \to \infty\}.\end{aligned}$$

A set $S$ is invariant if $\varphi_t(S) = S$ for all $t$, and forward invariant if $\varphi_t(S) \subseteq S$ for $t \ge 0$.

theorem

If the forward orbit of $\mathbf x_0$ stays in a compact $K \subset U$, then $\omega(\mathbf x_0)$ is non-empty, compact, connected and invariant, and $\operatorname{dist}(\varphi_t(\mathbf x_0), \omega(\mathbf x_0)) \to 0$.

full proof in the notes (N)

The proof is pure metric topology plus one input from unit 01. Non-emptiness and compactness are Bolzano–Weierstrass; invariance is continuous dependence on initial data:

$$\begin{aligned}\varphi_s(\mathbf y) &= \varphi_s\big(\lim \varphi_{t_n}(\mathbf x_0)\big)\\ &= \lim \varphi_{t_n + s}(\mathbf x_0) \in \omega(\mathbf x_0).\end{aligned}$$

why we want it

"Where does the orbit go?" becomes "what is $\omega(\mathbf x_0)$?", and that is a question about a set, which can be pinned down by inequalities. LaSalle (next) and Poincaré–Bendixson (section 4) are both statements about $\omega$-limit sets and nothing else.

Full proof

Non-empty and compact. Take $t_n = n$; the points $\varphi_n(\mathbf x_0)$ lie in the compact $K$, so a subsequence converges and its limit is in $\omega$. Each $\mathbf y \in \omega$ is a limit of points of $K$, so $\omega \subseteq K$ is bounded; and $\omega$ is closed, being an intersection: $\omega(\mathbf x_0) = \bigcap_{T \ge 0}\overline{\{\varphi_t(\mathbf x_0) : t \ge T\}}$, a nested intersection of non-empty compacta — which incidentally re-proves non-emptiness.

Invariant. Let $\mathbf y = \lim\varphi_{t_n}(\mathbf x_0)$ and fix $s$. Continuous dependence (unit 01, via Gronwall) says $\varphi_s$ is continuous where defined, so $\varphi_s(\mathbf y) = \lim \varphi_s(\varphi_{t_n}(\mathbf x_0)) = \lim\varphi_{t_n+s}(\mathbf x_0)$, and $t_n + s \to \infty$, so the limit lies in $\omega$. Applying this to $s$ and $-s$ gives $\varphi_s(\omega) = \omega$.

Convergence to $\omega$. If not, there are $\eta > 0$ and $t_n \to \infty$ with $\operatorname{dist}(\varphi_{t_n}(\mathbf x_0), \omega) \ge \eta$. Those points are in $K$, so a subsequence converges to some $\mathbf z$, which is in $\omega$ by definition and at distance $\ge \eta$ from $\omega$ — impossible.

Connected. If $\omega = A \sqcup B$ with $A, B$ non-empty, compact, and at positive distance $d$, then the orbit comes within $d/3$ of $A$ and of $B$ at arbitrarily large times, so by continuity of $t \mapsto \varphi_t$ and the intermediate value theorem applied to $t \mapsto \operatorname{dist}(\varphi_t, A)$ it is at distance exactly $d/2$ from $A$ at a sequence of times $s_n \to \infty$; a limit point of $\varphi_{s_n}$ lies in $\omega$ and in neither $A$ nor $B$. $\square$

Backward time gives $\alpha$-limit sets, with the same proofs. For the damped pendulum every $\omega$-limit set turns out to be a single equilibrium; for the systems of section 4 it can be a closed curve; in three dimensions it can be far stranger, which is unit 17.

02 · lyapunov

LaSalle's invariance principle

theorem (LaSalle, 1960)

Let $\Omega \subset U$ be compact and forward invariant, $V \in C^1$ with $\dot V \le 0$ on $\Omega$. Put

$$\begin{gathered}E = \{\mathbf x \in \Omega : \dot V(\mathbf x) = 0\},\\ M = \text{the largest invariant subset of } E .\end{gathered}$$

Then every solution starting in $\Omega$ satisfies $\operatorname{dist}(\varphi_t(\mathbf x_0), M) \to 0$.

proof

$V(\varphi_t)$ is non-increasing and bounded below on the compact $\Omega$, so $V(\varphi_t) \to c$. For $\mathbf y \in \omega(\mathbf x_0)$, continuity gives $V(\mathbf y) = c$: $V \equiv c$ on $\omega(\mathbf x_0)$. But $\omega$ is invariant, so the whole orbit through $\mathbf y$ lies in $\omega$ and $V$ is constant along it, whence $\dot V(\mathbf y) = 0$. Thus $\omega \subseteq E$ and $\omega$ is invariant, so $\omega \subseteq M$; and $\varphi_t \to \omega$. $\square$

what it fixes

Lyapunov's asymptotic half needs $\dot V < 0$ everywhere nearby. In mechanics that almost never holds: damping acts on velocity, so $\dot V$ vanishes on the whole set $\{\text{velocity} = 0\}$, not just at the equilibrium. LaSalle replaces "$\dot V < 0$" by "nothing can stay in $\dot V = 0$", which is exactly the true statement.

Note what it does not need: $V$ positive definite, or $\Omega$ small, or an equilibrium at all. It is a statement about where orbits go, and the conclusion is as global as $\Omega$ is large.

Finding $M$ in practice is a two-line argument every time: take a solution that lies in $E$ for all time, and push the constraint through the equations until it forces a point. For the damped pendulum, $E = \{\omega = 0\}$; a solution with $\omega \equiv 0$ has $\dot\omega \equiv 0$ too, and $\dot\omega = -\sin\theta$, so $\sin\theta = 0$. That is the whole calculation, and it is done on the next slide.

Why forward invariance of $\Omega$ is needed. Without it the orbit may leave $\Omega$, where nothing is assumed. In applications $\Omega$ is produced by the Lyapunov function itself: a bounded sublevel set $\{V \le c\}$ is forward invariant precisely because $\dot V \le 0$ on it. So one function supplies both hypotheses.

Attribution. J. P. LaSalle, "Some extensions of Liapunov's second method", IRE Transactions on Circuit Theory 7 (1960) 520–527. Barbashin and Krasovskii had the special case for globally positive definite $V$ a decade earlier. The principle is the reason control engineers can prove convergence for systems with no formula and plenty of damping, and it is the tool that makes the pendulum result below a theorem rather than a picture.

02 · lyapunov · example

The damped pendulum stops, and we can say where from

the theorem this unit was built for

Take $\gamma > 0$ and $\Omega = \{V \le c\}$ for any $c < 2$, inside the cylinder. Then:

$\Omega$ is compact and forward invariant. $\dot V = -\gamma\omega^2 \le 0$ keeps $V$ from rising; $V \le c < 2$ bounds $|\omega| \le \sqrt{2c}$ and confines $\theta$ to $|\theta| \le \arccos(1-c) < \pi$.
$E = \{\omega = 0\} \cap \Omega$, since $\dot V = 0$ iff $\omega = 0$.
$M = \{(0,0)\}$. A solution staying in $E$ has $\omega \equiv 0$, hence $\dot\omega \equiv 0$; but there $\dot\omega = -\sin\theta$, so $\sin\theta = 0$, and the only such point in $\Omega$ is the origin.

LaSalle: every orbit starting with $V < 2$ converges to the hanging position — letting $c \uparrow 2$, the basin contains all of $\{V < 2\}$.

what linearisation could not do

It gives asymptotic stability of the down position for $\gamma > 0$ — Hartman–Grobman applies, the equilibrium is hyperbolic — but only on some unspecified neighbourhood. The basin $\{V < 2\}$ is not a local statement and no linear argument produces it.

The compactness claim in detail. On the cylinder, $\{V \le c\}$ with $c < 2$ is $\{(\theta,\omega) : \tfrac12\omega^2 + 1 - \cos\theta \le c\}$. It requires $1 - \cos\theta \le c < 2$, i.e. $\cos\theta \ge 1 - c > -1$, so $|\theta|$ is bounded away from $\pi$; and $\omega^2 \le 2c$. Closed and bounded in the chart, hence compact. Forward invariance: if $V(\mathbf x_0) \le c$ then $V(\varphi_t(\mathbf x_0)) \le c$ because $V$ is non-increasing along orbits — and the orbit cannot leave the chart because it cannot cross $V = c$.

Why $c < 2$ and not $c = 2$. $\{V \le 2\}$ contains the saddle $(\pi, 0)$, so $M$ would contain it as well and the conclusion would weaken to "converges to the hanging position or to the upright one". Both do occur: the saddle's stable manifold — the orbits that run into it — is a one-dimensional set of initial conditions that really do end balanced. LaSalle is telling the exact truth again.

What happens outside $\{V < 2\}$. A fast enough initial spin makes the pendulum go over the top one or more times, losing energy each pass, until $V$ drops below $2$ — after which this theorem applies. Section 4 completes the argument by proving no periodic orbit exists anywhere, so no initial condition can rotate forever; every orbit of the damped pendulum ends at one of the two equilibria, and all but a one-dimensional set end hanging down.

02 · lyapunov · example

LaSalle's argument, drawn

$\gamma = \tfrac12$, $c = 1.5$. The orbit starts on the edge of $\Omega$ and never leaves it. It meets $E$ (circled) only by crossing it vertically — there $\dot\theta = 0$ but $\dot\omega = -\sin\theta \ne 0$ — so it cannot stay on $E$ anywhere except at $M$.

The three bullets of the previous slide, as a picture: $\Omega$ is the shaded oval, $E$ the orange segment of the axis, $M$ the single point at its centre.

checks

Limit $\gamma \to 0$: $E$ becomes all of $\Omega$ and $M$ all of it too — no convergence claimed, correctly.
Sharpness: $\{V < 2\}$ cannot be enlarged to $\{V \le 2\}$, which contains the saddle.
Numbers: released from rest at $\theta_0$, $V = 1-\cos\theta_0 < 2$ unless $\theta_0 = \pi$: from any angle but exactly upright, it ends hanging.

What to look for. Three things. The oval $\Omega = \{V \le 1.5\}$ reaches $|\theta| = \arccos(-0.5) = 2.094$ ($120°$) and $|\omega| = \sqrt3 = 1.732$, comfortably inside the dashed separatrix $V = 2$. The orbit, started on the boundary of $\Omega$ at $\theta = -1.6$, cuts inwards across every level curve of $V$ — it can never come back out. And each time it reaches the axis it goes straight through, in the vertical direction, because on the axis the field is $(0, -\sin\theta)$: this is the "push the constraint through the equations" step done by eye. The circled crossings are at $\theta = 0.78$, $-0.34$ and $0.15$, shrinking towards $M$.

Numbers for the release check: $V(\theta_0, 0) = 1 - \cos\theta_0$, which is $0.5$ at $60°$, $1$ at $90°$, $1.866$ at $150°$ and $2$ only at $180°$.

The orbit was integrated numerically, for the picture only; the theorem needed nothing from it. Compare "Lyapunov's proof, as a picture": there the same function $V$ only kept the orbit inside; here LaSalle uses the shape of $E$ to make it arrive.

section 4

Periodic orbits

Proving a cycle exists without computing it — and proving none does.

02 · cycles

Periodic orbits and limit cycles

definition

A periodic orbit is a non-equilibrium orbit with $\varphi_T(\mathbf x_0) = \mathbf x_0$ for some $T > 0$; the least such $T$ is its period. A limit cycle is a periodic orbit that is the $\omega$-limit set of some orbit not on it — an isolated cycle, which attracts or repels.

the distinction is not pedantic

The undamped pendulum has a continuum of periodic orbits and no limit cycle: each orbit is surrounded by others, none attracts. A limit cycle is a self-sustained oscillation with an amplitude of its own, picked out by the equations rather than by the initial data — a heartbeat, not a bell.

one you can see whole

$\dot x = x - y - x(x^2+y^2)$, $\dot y = x + y - y(x^2+y^2)$. With $r^2 = x^2+y^2$,

$$\begin{gathered}r\dot r = x\dot x + y\dot y = r^2 - r^4,\\ \dot\theta = (x\dot y - y\dot x)/r^2 = 1,\end{gathered}$$

so $\dot r = r(1-r^2)$ and $\dot\theta = 1$. The circle $r = 1$ is a periodic orbit of period $2\pi$; $r$ increases for $0 < r < 1$ and decreases for $r > 1$, so every orbit but the origin tends to it. A limit cycle, and we know it exactly because the system was built to be separable in polar form.

Almost nothing else is. The next slide is how one proves a cycle exists when nothing separates.

Checking the polar computation. $x\dot x + y\dot y = x(x - y - xr^2) + y(x + y - yr^2) = x^2 - xy - x^2r^2 + xy + y^2 - y^2r^2 = r^2 - r^4$; and $x\dot y - y\dot x = x(x+y-yr^2) - y(x-y-xr^2) = x^2 + xy - xyr^2 - xy + y^2 + xyr^2 = r^2$. Both cross terms cancel, which is the design of the example.

Why the period is exactly $2\pi$. On the cycle $r \equiv 1$ and $\dot\theta = 1$, so the angle takes time $2\pi$ to advance by $2\pi$. In the units of a mechanical problem that $1$ would be a frequency and the period $2\pi/\omega_0$; dimensionally the period of any cycle is (angle)/(angular velocity), a time, as required.

An orbit that returns is periodic. If $\varphi_T(\mathbf x_0) = \mathbf x_0$ for some $T > 0$ then by the group property $\varphi_{t+T} = \varphi_t$ for all $t$, so the whole solution is $T$-periodic — there is no such thing as a solution that comes back once. The set of such $T$ is a closed subgroup of $(\mathbb{R}, +)$, hence either $\{0\}$, all of $\mathbb{R}$ (an equilibrium) or $T_0\mathbb{Z}$ with $T_0$ the minimal period. That is elementary group theory doing a small honest job.

Limit cycles are what makes clocks, lasers, heartbeats and the Van der Pol circuit of problem P6 possible: the amplitude is a property of the system, so it is reproducible and robust to a knock. A conservative oscillator cannot do this — knock it and the amplitude changes forever.

02 · cycles

Poincaré–Bendixson

theorem (Poincaré 1885, Bendixson 1901) · cited

Let $f$ be $C^1$ on an open $U \subseteq \mathbb{R}^2$, and suppose the forward orbit of $\mathbf x_0$ stays in a compact $K \subset U$ containing no equilibrium. Then $\omega(\mathbf x_0)$ is a periodic orbit.

not proved here: Perko §3.7, or Hartman, Ordinary Differential Equations, ch. VII

In practice one supplies $K$ as a trapping region: a compact set with the field pointing inwards along its boundary, so no orbit can leave. If it also contains no equilibrium, a cycle exists inside it — proved, without any idea what the cycle looks like.

Why only the plane? The next slide draws the step that needs it.

it is a theorem about the plane

"$C$ has an inside, and getting in means crossing $C$" is the Jordan curve theorem. In $\mathbb{R}^3$ a closed curve has no inside, the argument dies, and the conclusion is false — a bounded, equilibrium-free orbit can be chaotic. That is unit 17.

So two-dimensionality is not a convenience of the proof. It is the content.

The proof, in outline

A transverse segment $\Sigma$ is a closed line segment on which $f$ is nowhere tangent, so every orbit crosses it in the same direction. Step 1: successive intersections $\mathbf p_1, \mathbf p_2, \dots$ of one orbit with $\Sigma$ are monotone along $\Sigma$. Proof: the arc of orbit from $\mathbf p_1$ to $\mathbf p_2$ together with the piece of $\Sigma$ between them is a simple closed curve; by the Jordan curve theorem it splits the plane, and the flow crosses $\Sigma$ one way only, so the orbit is confined to one side afterwards and cannot return past $\mathbf p_1$. Step 2: an $\omega$-limit set meets a transverse segment in at most one point — otherwise monotonicity is violated. Step 3: take $\mathbf y \in \omega(\mathbf x_0)$; $\omega$ is invariant and compact, so $\omega(\mathbf y) \subseteq \omega(\mathbf x_0)$ is non-empty; pick $\mathbf z$ in it, which is not an equilibrium by hypothesis, and lay a transverse segment through $\mathbf z$. The orbit of $\mathbf y$ meets it infinitely often but $\omega(\mathbf x_0)$ meets it once, so the orbit of $\mathbf y$ returns to the same point: it is periodic. A last step shows $\omega(\mathbf x_0)$ is that whole orbit and nothing more.

Everything above except the Jordan curve theorem is elementary; the Jordan curve theorem is exactly the input that has no analogue in higher dimensions. The theorem is stated as cited here because doing step 1 honestly needs the flow-box theorem (local rectification of a vector field), which needs the inverse function theorem of unit 03.

The general form allows equilibria: if $K$ contains finitely many, then $\omega(\mathbf x_0)$ is an equilibrium, a periodic orbit, or a union of equilibria and orbits joining them (a "graphic", such as a homoclinic loop). The undamped pendulum's separatrix is exactly such a union, which is why it is worth knowing the general statement.

02 · cycles

Why the plane: crossings of a transverse segment

$\Sigma$ is a segment the field crosses in one direction only. Follow one orbit from its crossing $p_1$ round to its next crossing $p_2$; that arc plus the piece of $\Sigma$ from $p_2$ back to $p_1$ is a closed curve $C$.

The orbit cannot re-cross its own arc (orbits do not cross), and it crosses the piece of $\Sigma$ only outwards. So after $p_2$ it is shut out of $C$ for good, and $p_3$ lies beyond $p_2$: the crossings are monotone along $\Sigma$.

$\dot r = 0.15\,r(1-r^2)$, $\dot\theta = 1$, from $r = 0.2$ on $\Sigma$. Thick: the first lap, which with the segment from $p_2$ to $p_1$ bounds the shaded inside of $C$. The crossings sit at $r = 0.200,\ 0.464,\ 0.802,\ 0.960$ — monotone, as forced.

What to look for. The shaded region is the inside of $C$. Along the segment piece from $p_1$ to $p_2$ the flow goes from below the axis to above it, which for this $C$ means from the inside to the outside. So orbits can leave the shaded region through that piece but never enter it, and they cannot pass through the thick arc at all. The orbit, once past $p_2$, is outside for ever — and its next return to $\Sigma$ must land outside the piece $[p_1, p_2]$, which on this side means further out. That is step 1 of the proof outline in the notes of "Poincaré–Bendixson".

Where the plane is used. "$C$ has an inside and an outside, and to get from one to the other you must cross $C$" is the Jordan curve theorem. In three dimensions a closed curve separates nothing: the orbit can simply go round it, so the crossings of a transverse disc need not be monotone, and they are not in the chaotic systems of unit 17.

The system and the numbers. This is the limit-cycle example of "Periodic orbits and limit cycles", slowed by the factor $0.15$ so that the approach to $r = 1$ takes several turns. Its exact solution is $r(t) = r_0 e^{0.15t}/\sqrt{1 + r_0^2(e^{0.3t}-1)}$ (separation of variables, unit 00), and the crossings are $r(2\pi k)$ for $k = 0, 1, 2, 3$: $0.2000$, $0.4640$, $0.8024$, $0.9605$, then $0.9937$, converging on the cycle. In this example we could compute them; the point of step 1 is that their order is forced even when we cannot.

02 · cycles · example

A trapping annulus, with the inequalities shown

the construction

For $\dot r = r(1-r^2)$, $\dot\theta = 1$ take $K = \{\tfrac12 \le r \le 2\}$, compact.

On $r = \tfrac12$: $\dot r = \tfrac12(1-\tfrac14) = +0.375 > 0$, pointing outwards into $K$.
On $r = 2$: $\dot r = 2(1-4) = -6 < 0$, pointing inwards into $K$.
The only equilibrium of the system is the origin ($\dot\theta = 1 \ne 0$ elsewhere), and it is not in $K$.

Poincaré–Bendixson: every orbit starting in $K$ has a periodic orbit as its $\omega$-limit set, so a cycle exists in $K$. Here we happen to know it is $r=1$ with period $2\pi$; the argument never used that.

Problem P6 runs the same three steps on Van der Pol, where no formula for the cycle exists at all.

The annulus $K$, shaded. The orange arrows are the field on its two edges: outwards-and-round at $r = \tfrac12$, inwards-and-round at $r = 2$ — into $K$ both times. One orbit from each edge, from the exact solution, winds onto $r = 1$ within a turn.

Why "pointing outwards at $r = \tfrac12$" is inwards for $K$. $K$ is the annulus, so its boundary has two pieces and "into $K$" means away from the hole at the inner circle and towards the centre at the outer one. Both inequalities say the same thing: once in $K$, always in $K$. A trapping region need not be simply connected, and for Poincaré–Bendixson it usually must not be — a simply connected trapping region with no equilibrium is impossible. Why: along the boundary of a disc an inward-pointing field is homotopic, without vanishing, to the pure inward normal $-\boldsymbol\xi$, whose index is $+1$; the index theorem of section 5 then demands equilibria inside summing to $+1$, and there are none.

Forward invariance from a strict inequality. If $g \in C^1$, $K = \{g \le 0\}$ is compact and $\dot g < 0$ on $\{g = 0\}$, then no orbit leaves $K$: at a first exit time $t_1$ we would have $g(\mathbf x(t_1)) = 0$ and $g > 0$ just after, so $\dot g(\mathbf x(t_1)) \ge 0$, a contradiction. Here $g = \tfrac14 - r^2$ and $g = r^2 - 4$, and $\dot g$ is $-2r\dot r$ and $2r\dot r$ respectively; both are negative on their boundary pieces. When the inequality is only $\le$, the same conclusion needs a little more care, and problem P6 shows how to handle the two tangential points it meets.

The explicit solution, for comparison only. $\dot r = r(1-r^2)$ separates (unit 00): $r(t) = r_0e^t/\sqrt{1 + r_0^2(e^{2t}-1)} \to 1$ for every $r_0 > 0$. The drawn spirals use it. But an example one can solve is a bad advertisement for a theorem one needs when one cannot, so read the box, not the picture.

02 · cycles

Ruling cycles out

theorem · the damped pendulum has no periodic orbit

Let $\gamma > 0$ and suppose $\varphi_t$ is $T$-periodic. Then

$$\begin{aligned}0 &= V(\mathbf x(T)) - V(\mathbf x(0))\\ &= \int_0^T \dot V\,\mathrm{d}t = -\gamma\int_0^T \omega^2\,\mathrm{d}t \le 0,\end{aligned}$$

so $\omega \equiv 0$ on the orbit, hence $\dot\omega = -\sin\theta \equiv 0$, and the orbit is one of the two equilibria — not a periodic orbit. $\square$

Three lines, valid on the whole cylinder, covering both the swinging orbits and the rotating ones. It needs no two-dimensionality and no index theory: a strictly decreasing quantity is incompatible with returning.

the general tool · cited forward

Bendixson–Dulac criterion. If $\partial_x f_1 + \partial_y f_2$ has one strict sign on a simply connected region $D$ — one with no holes, so that every closed curve in it can be shrunk to a point inside it — then $D$ contains no periodic orbit.

Proof by Green's theorem: $\oint_\Gamma (f_1\,\mathrm{d}y - f_2\,\mathrm{d}x) = 0$ along an orbit, while the double integral of the divergence over the enclosed region is nonzero. The line and area integrals arrive in units 04 and 05, and the proof is given there.

For the pendulum $\partial_\theta(\omega) + \partial_\omega(-\gamma\omega-\sin\theta) = -\gamma < 0$: the criterion applies at once. Note it needs simply connected, so on the cylinder it kills only the contractible cycles — the energy argument above is the one that also kills rotation.

Why the energy argument is worth more than the criterion here. The cylinder is not simply connected. A rotating pendulum orbit goes once round it and bounds no disc, so Bendixson–Dulac says nothing about it, and neither will the index theorem of section 5. The integral argument does not care: it only uses that $V$ is a well-defined function on the cylinder that strictly decreases whenever the pendulum moves.

Dulac's refinement replaces $f$ by $\rho f$ for a positive $C^1$ function $\rho$, which does not change the orbits but can change the sign of the divergence. Finding a suitable $\rho$ is as much of an art as finding a Lyapunov function, and for the same reason: both are certificates, easy to check and hard to invent.

The full picture of the damped pendulum, now complete. Every forward orbit is bounded (the note on "The flow is a one-parameter group"), so it has a non-empty compact invariant $\omega$-limit set. That set contains no periodic orbit by the theorem above. Poincaré–Bendixson in the general form then forces it to be an equilibrium or a union of equilibria and connecting orbits — and since $V$ is constant on an $\omega$-limit set while $\dot V < 0$ off $\omega = 0$, it must be a single equilibrium. So every orbit converges to the hanging or the upright position. "The damped pendulum stops, and we can say where from" gave the basin of the first; this gives the dichotomy.

section 5

The index of a vector field

An integer attached to a closed curve, which counts what is inside.

02 · index

The index of a closed curve

definition

Let $f$ be a continuous planar field and $\Gamma$ a closed curve $\sigma : [0,1] \to \mathbb{R}^2$, $\sigma(0)=\sigma(1)$, on which $f \ne 0$. There is a continuous $\vartheta : [0,1] \to \mathbb{R}$ with

$$f(\sigma(s)) = |f(\sigma(s))|\,\big(\cos\vartheta(s),\, \sin\vartheta(s)\big),$$

unique up to a constant in $2\pi\mathbb{Z}$. The index of $\Gamma$ is

$$I_f(\Gamma) = \frac{\vartheta(1) - \vartheta(0)}{2\pi} \in \mathbb{Z}.$$

In words: the net number of turns the field makes as you walk once round $\Gamma$ — read off the arrows, with no derivatives, no solving and no linearity.

theorem · homotopy invariance

If $\Gamma_0$ and $\Gamma_1$ are homotopic through closed curves on which $f$ never vanishes — joined by a continuous family $\Gamma_u$, $u \in [0,1]$, of such curves — then $I_f(\Gamma_0) = I_f(\Gamma_1)$.

Why: the index along the homotopy is a continuous $\mathbb{Z}$-valued function of the parameter, hence constant on $[0,1]$.

two immediate corollaries

1. A curve shrinkable to a point inside a region where $f \ne 0$ has index $0$: a constant curve turns not at all.
2. No deformation that avoids the zeros of $f$ changes the index — it depends only on which equilibria are enclosed.

Full proof that the continuous angle exists

The map $g = f\circ\sigma / |f\circ\sigma| : [0,1] \to S^1$ is continuous on a compact interval, hence uniformly continuous: choose a partition $0 = s_0 < \dots < s_N = 1$ so fine that $|g(s) - g(s_i)| < 1$ for $s \in [s_i, s_{i+1}]$, which keeps $g$ on that subinterval inside an open half-circle. On a half-circle the angle has a continuous branch (an explicit arctangent), so define $\vartheta$ on $[s_0,s_1]$ by that branch, then on $[s_1,s_2]$ by the branch agreeing with $\vartheta(s_1)$, and so on. The result is continuous, and any two such differ by a continuous integer multiple of $2\pi$, hence by a constant. Since $g(1) = g(0)$, $\vartheta(1) - \vartheta(0) \in 2\pi\mathbb{Z}$, so the index is an integer. $\square$ (This is the covering-space lifting property of $\mathbb{R} \to S^1$, proved by hand.)

Why the index varies continuously with the homotopy. Let $H : [0,1]^2 \to \mathbb{R}^2$ be the homotopy, with $f\circ H \ne 0$. $|f \circ H|$ has a positive minimum by compactness, and $f\circ H$ is uniformly continuous, so the same partition argument lifts $H$ to a continuous angle $\vartheta(s, u)$; then $u \mapsto \vartheta(1,u) - \vartheta(0,u)$ is continuous and $2\pi\mathbb{Z}$-valued. $\square$

The index is the winding number of $f$ along $\Gamma$ about the origin of the target plane, not of $\Gamma$ about a point of the domain. Confusing the two is the standard first error. It is also the degree of a map $S^1 \to S^1$, and this whole slide is degree theory in its smallest non-trivial case — the case that, in unit 34's language, will be an integral of a $1$-form.

02 · index · example

Reading an index off the arrows

Two circles $\Gamma_1, \Gamma_2$ of radius $1.25$ in the pendulum's plane, round $(0,0)$ and $(\pi,0)$, with the field drawn at sixteen points of each. Right: the continuous angle $\vartheta(s)$ of the field as each circle is walked once anticlockwise.

Round $(0,0)$ the arrow turns once with the walk: $\vartheta$ gains $2\pi$, index $+1$. Round $(\pi, 0)$ it turns once against it: $\vartheta$ loses $2\pi$, index $-1$. Nothing was solved — the arrows are $f$ evaluated on the circle, $\gamma = \tfrac12$.

How to read the arrows by hand. Pick a starting point on $\Gamma_1$, say its rightmost point $(1.25, 0)$: there $\omega = 0$ and the field is $(0, -\sin 1.25)$, pointing straight down. Walk anticlockwise to the top, $(0, 1.25)$: the field is $(1.25, -0.625)$, pointing right and a little down. At the leftmost point it points straight up, at the bottom left and a little up. Down, right, up, left: one full anticlockwise turn while you walked one anticlockwise turn. Do the same on $\Gamma_2$ and the arrows turn the other way round. The right-hand plot is that bookkeeping done continuously, which is exactly the lift $\vartheta$ of the definition.

Why the plot ends at integers. The walk returns to its start, so the arrow returns to its starting direction, and $\vartheta(1) - \vartheta(0)$ must be a multiple of $2\pi$. The plot is computed: $\vartheta$ is the angle of $f$ at $721$ points of each circle, unwrapped by adding the multiple of $2\pi$ that keeps each step below $\pi$ in size — the partition argument of the previous slide's notes, done numerically. The end values are $+1$ and $-1$ to rounding error.

Homotopy invariance, seen. Shrink $\Gamma_1$ towards $(0,0)$ or swell it to any radius below $\pi$: no zero of $f$ is crossed, so the curve on the right deforms but its end value cannot move off $+1$. Swell it past $\pi$ and it swallows the upright points, whose indices then count too.

02 · index

Indices of equilibria, and what they count

For an isolated equilibrium $\mathbf x^*$, the index of an equilibrium $\operatorname{ind}(\mathbf x^*)$ is $I_f(\Gamma)$ for any small circle around it; homotopy invariance makes the choice irrelevant.

theorem · an invertible $Df$ settles it

If $A = Df(\mathbf x^*)$ is invertible then $\operatorname{ind}(\mathbf x^*) = \operatorname{sign}(\det A)$: $\;+1$ for nodes, spirals, stars and centres, $\;-1$ for saddles. Hyperbolic is not needed — only $\det A \ne 0$, which is why centres are included.

proof in the notes (N)

theorem · the Poincaré index theorem

If $\Gamma$ bounds a disc $D$ with $f \ne 0$ on $\Gamma$ and finitely many equilibria inside, $\;I_f(\Gamma) = \sum_{\mathbf x^* \in D}\operatorname{ind}(\mathbf x^*)$.

theorem (Umlaufsatz, Hopf 1935) · cited

A periodic orbit has index $+1$: the field is tangent to the curve, and the tangent of a simple closed curve turns once.

not proved here: do Carmo, Differential Geometry of Curves and Surfaces, §5-7

Walk anticlockwise round each circle and watch the arrow. On the left it turns once anticlockwise; on the right, once clockwise.

Proof of the index theorem. Deform $\Gamma$ inside $D$ into a "flower": small circles $\Gamma_i$ round each equilibrium joined by pairs of thin corridors traversed in opposite directions. The corridor contributions cancel, and the deformation never crosses a zero, so $I_f(\Gamma) = \sum_i I_f(\Gamma_i) = \sum_i \operatorname{ind}(\mathbf x_i^*)$. The one step not done here is that $D$ minus finitely many points deformation-retracts onto such a flower; that is standard plane topology, and the whole construction is written out in Perko's chapter on index theory in the plane (ch. 3).

Full proof that $\operatorname{ind}(\mathbf x^*) = \operatorname{sign}\det A$

Take $\mathbf x^*$ at the origin. On a circle $|\boldsymbol\xi| = \rho$ we have $|A\boldsymbol\xi| \ge c\rho$ with $c = 1/\|A^{-1}\| > 0$, while $|f(\boldsymbol\xi) - A\boldsymbol\xi| = o(\rho)$; so for $\rho$ small enough the straight-line homotopy $A\boldsymbol\xi + u\,(f(\boldsymbol\xi) - A\boldsymbol\xi)$, $u \in [0,1]$, never vanishes on that circle, and homotopy invariance gives $\operatorname{ind}(\mathbf x^*) = I_A(\Gamma_\rho)$. It remains to compute the index of a linear field. For the sink $A = -I$, on $\sigma(s) = \rho(\cos 2\pi s, \sin 2\pi s)$ the field has angle $2\pi s + \pi$, so $\vartheta$ increases by $2\pi$: index $+1$. For the saddle $A = \operatorname{diag}(1,-1)$ the field at angle $\phi$ has angle $-\phi$, decreasing by $2\pi$: index $-1$. Every other invertible $A$ is joined to one of these two by a path inside $GL_2(\mathbb{R})$, along which the index is a continuous integer, hence constant. $\square$

Why such a path exists. $\det$ is continuous and non-zero on $GL_2(\mathbb{R})$, so it keeps its sign along any path: the two determinant signs are in different components. Conversely Gaussian elimination writes any $A$ as a product of elementary matrices, and each of those is joined to $I$ by scaling its off-diagonal entry to $0$ or its diagonal entry continuously to $\pm1$ without passing through a singular matrix; the resulting path carries $A$ to $I$ or to $\operatorname{diag}(1,-1)$. So $GL_2(\mathbb{R})$ has exactly two connected components, and they are the two determinant signs.

The consequences are what we actually use. A periodic orbit has index $+1$, so the equilibria it encloses have indices summing to $+1$. Therefore: (i) a periodic orbit must enclose at least one equilibrium; (ii) a region with no equilibria contains no periodic orbit; (iii) a region whose only equilibrium is a saddle contains none either, since $-1 \ne +1$. Each is a genuine non-existence theorem obtained by counting.

The same counting on a surface gives the Poincaré–Hopf theorem, cited: for a compact surface, with the field pointing inwards along any boundary, the indices of its isolated zeros sum to the Euler characteristic — $2$ for a sphere (hence the hairy ball theorem), $0$ for a torus, and $0$ for a compact piece of cylinder, which is a sphere with two discs removed. Proof: Milnor, Topology from the Differentiable Viewpoint, §6. That is why the pendulum's $+1$ and $-1$ cancel.

02 · index · example

The pendulum: $+1$, $-1$, and the orbits that escape the theorem

counting on the cylinder

$\det Df = \cos\theta$ at an equilibrium, so $(0,0)$ has index $\operatorname{sign}(+1) = +1$ and $(\pi,0)$ has $\operatorname{sign}(-1) = -1$: the two indices sum to zero, and along the line of equilibria $(k\pi, 0)$ in the covering strip they alternate $+1, -1, +1, \dots$

Consequences, for any $\gamma$ including $0$:

a contractible periodic orbit must enclose equilibria of total index $+1$, so it must enclose the hanging position — and cannot enclose the upright one alone;
any region of the cylinder containing no equilibrium, or only the saddle, contains no contractible cycle at all.

the escape clause, and it is real

A rotating pendulum orbit — the bob going over the top again and again, which exists for $\gamma = 0$ and large enough energy — is a closed orbit encircling no equilibrium. It does not contradict anything: it wraps once round the cylinder, so it bounds no disc, and the index theorem assumes a disc.

the running example, closed for cycles

$\gamma = 0$: contractible cycles exist (the swinging orbits inside the separatrix, each enclosing the index-$+1$ point, as required), and non-contractible ones exist (rotations).
$\gamma > 0$: neither exists — the energy argument of "Ruling cycles out", which never mentions the index.

Why $\det Df = \cos\theta$. $Df = \begin{pmatrix}0&1\\-\cos\theta&-\gamma\end{pmatrix}$ has determinant $0\cdot(-\gamma) - 1\cdot(-\cos\theta) = \cos\theta$. At $\theta = 0$ that is $+1$ and at $\theta = \pi$ it is $-1$, for every $\gamma$ — the damping never enters the index, which is the point: the index is a robust, discrete invariant, blind to everything the parameter does until an equilibrium actually collides with another. Section 6 is about exactly that moment.

The sum is zero, and that is forced. The $\pm1$ above was computed directly, for every $\gamma$; here is why it could not have come out otherwise. Take $\gamma > 0$ and cut the cylinder at $|\omega| = \Omega$ with $\Omega > 1/\gamma$: there $\dot\omega = -\gamma\omega - \sin\theta$ has the sign opposite to $\omega$, so the field points strictly inwards along both cut circles. Poincaré–Hopf then makes the total index the Euler characteristic of that compact cylinder, which is zero. So equilibria must come in $\pm$ pairs. The pendulum has one of each, and a pendulum with a torque (section 6) loses both at once, never one.

How to use this in practice. Given a planar system, find the equilibria, compute $\operatorname{sign}\det Df$ at each, and you have ruled out cycles in large parts of the plane before doing anything else. It costs two determinants and it is the cheapest global information available anywhere in this unit.

section 6

Bifurcations

What the portrait does as you turn a knob.

02 · bifurcations

Families, and where the portrait breaks

definition

A family $\dot{\mathbf x} = f(\mathbf x, \mu)$ has a bifurcation at $\mu_0$ if the phase portrait at $\mu_0$ is not topologically equivalent to those at nearby $\mu$ — no homeomorphism carries its orbits onto theirs, keeping the direction of time (the speed along orbits may change, unlike a conjugacy): equilibria or cycles are created, destroyed or change type.

By Hartman–Grobman, nothing can happen while every equilibrium is hyperbolic and stays so. Bifurcations live exactly where hyperbolicity fails — on the two lines $\Delta = 0$ and $\tau = 0$ of "Planar linear systems: two numbers decide". That is the whole organising principle.

the saddle-node bifurcation

Normal form $\dot x = \mu - x^2$. For $\mu > 0$ two equilibria $x = \pm\sqrt\mu$, one stable ($f'=-2\sqrt\mu<0$) and one unstable; at $\mu = 0$ they collide at $x=0$ with $f'(0) = 0$; for $\mu < 0$ none at all.

Solid: stable; dashed: unstable. The dotted verticals are phase lines, arrows giving the sign of $\dot x = \mu - x^2$: at $\mu < 0$ every $x$ falls, at $\mu = 0.64$ the flow runs into $x = 0.8$ and away from $x = -0.8$.

Why "normal form". The claim behind the name is that any one-parameter family whose equilibrium has $f = 0$, $\partial_x f = 0$, $\partial_{xx} f \ne 0$ and $\partial_\mu f \ne 0$ at the bifurcation point is, after a smooth change of coordinates and of parameter, equal to $\dot x = \mu - x^2$ near that point. So the picture drawn once is the picture in every such case. Proving it is the implicit function theorem plus Taylor expansion — unit 03 supplies the first, and the statement is used here as a classification, with the verification done case by case on the examples.

Where the name comes from. In the plane, the two colliding equilibria are a saddle and a node: the two-dimensional version is $\dot x = \mu - x^2$, $\dot y = -y$, whose equilibria have Jacobians $\operatorname{diag}(\mp2\sqrt\mu, -1)$ — a stable node and a saddle. Their determinants are $\pm2\sqrt\mu$, so their indices are $+1$ and $-1$: a saddle-node bifurcation annihilates a $+1$ with a $-1$, and the total index is conserved, exactly as the previous slide's counting demands. That is not a coincidence, it is a theorem about degree.

Reading the diagram. Horizontal axis the parameter, vertical the equilibrium position, solid for stable and dashed for unstable. It is a picture of a set — the zero set of $f(x,\mu)$ — not of a trajectory, and it is the standard way to summarise a whole family at once. Every bifurcation diagram in physics, up to and including phase transitions, is read the same way.

02 · bifurcations

Transcritical and pitchfork

the transcritical bifurcation

Normal form $\dot x = \mu x - x^2$. Equilibria $x = 0$ and $x = \mu$ for every $\mu$: they cross at $\mu = 0$ and exchange stability, since $f'(0) = \mu$ and $f'(\mu) = -\mu$. Nothing is created or destroyed.

the pitchfork bifurcation

Supercritical normal form $\dot x = \mu x - x^3$: $x=0$ for all $\mu$, plus $x = \pm\sqrt\mu$ for $\mu > 0$, which are stable ($f' = -2\mu$) while $x=0$ turns unstable.

Subcritical $\dot x = \mu x + x^3$: the two extra branches exist for $\mu < 0$ and are unstable. Same picture, reflected — and physically opposite, since nothing nearby catches the system when $x=0$ loses stability.

Solid stable, dashed unstable. Transcritical: $x = 0$ and $x = \mu$ swap stability where they cross. Supercritical pitchfork: as $x = 0$ turns unstable, the stable pair $x = \pm\sqrt\mu$ grows out of it.

When each one occurs. A transcritical bifurcation needs a solution branch that persists for all $\mu$ — typically $x = 0$ forced by the physics, as in a population that can always be zero or a laser below threshold. A pitchfork needs a symmetry: if $f(-x,\mu) = -f(x,\mu)$, then equilibria come in $\pm$ pairs and a single branch cannot peel off on its own. Buckling of a symmetric strut is the mechanical example, and the two new equilibria are "buckled left" and "buckled right". Take the symmetry away and the pitchfork breaks into a saddle-node plus a surviving branch — the imperfect bifurcation, which problem P9 does not ask for but which is the reason real struts always buckle the same way.

Supercritical versus subcritical matters more than the names suggest. Supercritical: as $\mu$ crosses zero the state moves continuously to a nearby new equilibrium, a soft transition. Subcritical: at $\mu = 0$ the state has nowhere nearby to go and leaves the neighbourhood entirely — a jump, hysteresis, and in engineering a failure. The algebraic difference is one sign in a cubic term.

Stability by the derivative. In one dimension the whole classification is $\operatorname{sign} f'(x^*)$, which is Hartman–Grobman in its simplest case: $f' < 0$ stable, $f' > 0$ unstable, $f' = 0$ undecided and therefore exactly where the bifurcations sit. Each normal form above has $f' = 0$ at its bifurcation point, as it must.

02 · bifurcations

Hopf: an equilibrium trades stability for a cycle

the Hopf bifurcation, normal form

$\dot r = \mu r - r^3$, $\dot\theta = \omega_0$, i.e.

$$\begin{aligned}\dot x &= \mu x - \omega_0 y - x(x^2+y^2),\\ \dot y &= \omega_0 x + \mu y - y(x^2+y^2).\end{aligned}$$

Eigenvalues $\mu \pm i\omega_0$ at the origin: stable spiral for $\mu < 0$, unstable for $\mu > 0$, non-hyperbolic exactly at $\mu = 0$. For $\mu > 0$ a limit cycle appears at $r = \sqrt\mu$, period $2\pi/\omega_0$.

the theorem · cited

If a planar family has eigenvalues $\alpha(\mu) \pm i\beta(\mu)$ crossing the imaginary axis transversally ($\alpha(0)=0$, $\alpha'(0) \ne 0$, $\beta(0) \ne 0$) and one cubic coefficient is non-zero, a unique small limit cycle is born, amplitude $\propto\sqrt{|\mu|}$, period $\to 2\pi/\beta(0)$.

Marsden & McCracken, The Hopf Bifurcation and its Applications, §3; Perko §4.4

two checks on the normal form

Amplitude: $\dot r = 0$ at $r = \sqrt\mu$; $\partial_r(\mu r - r^3) = \mu - 3r^2 = -2\mu < 0$ there, so the cycle attracts. At $\mu = 0.25$ the amplitude is exactly $0.5$.
Scaling: $r \sim \sqrt\mu$ means the cycle grows as the square root of the distance past the threshold — the signature that distinguishes a Hopf bifurcation in data from anything else.

The next slide draws both sides of the threshold.

"When linearisation lies" was already this. The family $\dot r = ar^3$, $\dot\theta = 1$ is the degenerate Hopf normal form with $\mu \equiv 0$: the linear term never crosses, and the cubic alone decides. Adding $\mu r$ to it produces exactly the family above, so the example that broke linearisation and the bifurcation that organises the breaking are the same two lines of algebra.

Why the cubic coefficient must be checked. The linear crossing tells you the equilibrium changes stability; it does not tell you what catches the orbit. The sign of the first Lyapunov coefficient (the cubic term after normal-form reduction) decides supercritical versus subcritical, and computing it for a real system is a genuinely messy calculation — which is why the theorem is cited and its hypotheses are stated carefully rather than waved at.

Physically, a Hopf bifurcation is the birth of an oscillation from a steady state: the onset of flutter, the Belousov–Zhabotinsky reaction starting to tick, a laser passing threshold, the Van der Pol circuit of problem P6 as its feedback is turned up. In every case the amplitude is not free, unlike a conservative oscillator's, and the $\sqrt{\mu}$ growth is measurable.

There is no Hopf bifurcation in one dimension: complex eigenvalues need two. This is the first bifurcation on the deck that is genuinely two-dimensional, and the reason section 6 is placed after section 4 rather than before.

02 · bifurcations · example

Hopf, drawn on both sides of the threshold

The normal form with $\omega_0 = 1$. Left: a stable spiral. Middle: the origin has turned unstable, and orbits from inside and outside both wind onto the cycle $r = \sqrt{0.25} = 0.5$. Right: the whole family at once — the cycle is born at $\mu = 0$ with zero radius and grows like $\sqrt\mu$.

subcritical again

With $+r^3$ instead, the cycle exists for $\mu < 0$ and is unstable: it is the boundary of the basin of the stable origin, and it shrinks to nothing as $\mu \to 0^-$. Past the threshold the system jumps somewhere far away. Aeroelastic flutter is this case, and it is why the distinction is worth a slide.

What to look for. Compare the two portraits near the origin: at $\mu = -0.25$ the spiral tightens by $e^{-0.25 \cdot 2\pi} = 0.21$ per turn; at $\mu = +0.25$ it opens by $e^{0.25\cdot 2\pi} = 4.8$ per turn near the origin, but the cubic term catches it and the orbit settles on $r = 0.5$. Nothing is created discontinuously: the cycle starts at radius $0$, which is what "supercritical" means and why the onset is gentle. The right-hand panel is the bifurcation diagram, drawn the same way as those of the previous two slides — position of the invariant set (here the cycle's radius) against the parameter.

The numbers. On $\dot r = \mu r - r^3$ the cycle is $r = \sqrt\mu$; at $\mu = 0.25$ that is $0.5$, as the previous slide's check computed. The portraits were integrated by Runge–Kutta from $r = 1.1$ and (middle) $r = 0.04$; the circle is drawn exactly, not from the integration.

Subcritical, in the same picture. Flip the sign of the cubic term and the right-hand panel reflects: the $\sqrt{-\mu}$ branch lies over $\mu < 0$ and is unstable. The middle portrait would then have no cycle to catch the orbit, which runs away — the jump the box describes.

02 · bifurcations · example

The pendulum with a constant torque

the running example, with a knob

Drive it: $\ddot\theta + \gamma\dot\theta + \sin\theta = T$, $T$ the applied torque in units of $mgl$. Equilibria need $\omega = 0$ and $\sin\theta = T$:

$T < 1$: two on the cylinder, $\theta_1 = \arcsin T$ (with $\cos\theta_1 > 0$) and $\theta_2 = \pi - \arcsin T$ ($\cos\theta_2 < 0$);
$T = 1$: one, at $\theta = \pi/2$, with $\det Df = \cos\theta = 0$ — non-hyperbolic;
$T > 1$: none — the torque beats gravity at every angle and the pendulum turns forever.

$\det Df = \cos\theta$ and $\operatorname{tr}Df = -\gamma$, so $\theta_1$ is a sink and $\theta_2$ a saddle: a saddle-node bifurcation at $T = 1$, a $+1$ and a $-1$ annihilating as the note on "Families, and where the portrait breaks" requires.

numbers, at $\gamma = \tfrac12$, $T = 0.8$

Sink at $\theta_1 = 0.9273$ rad $= 53.13°$, $\cos\theta_1 = 0.6$: eigenvalues $-0.25 \pm 0.7331\,i$ — still a spiral, but slower than the undriven $0.9682$.
Saddle at $\theta_2 = 2.2143$ rad $= 126.87°$, $\cos\theta_2 = -0.6$: eigenvalues $0.5639$ and $-1.0639$.

Every number. $\arcsin 0.8 = 0.927295$ rad $= 53.1301°$; $\cos = 0.6$. Eigenvalues of $\begin{pmatrix}0&1\\-0.6&-0.5\end{pmatrix}$: $\lambda = \tfrac12(-0.5\pm\sqrt{0.25-2.4})$, complex with $\beta = \tfrac12\sqrt{2.15} = 0.733144$. At $\theta_2$: $\cos = -0.6$, $\lambda = \tfrac12(-0.5\pm\sqrt{0.25+2.4}) = \tfrac12(-0.5 \pm 1.627882)$, giving $0.563941$ and $-1.063941$; product $-0.600000 = \det$, sum $-0.5 = \tau$. Both checks pass. The spiral-to-node threshold: $4\cos\theta_1 = \gamma^2 = 0.25$ gives $\cos\theta_1 = 0.0625$ and $T = \sin\theta_1 = \sqrt{1 - 0.0625^2} = 0.998045$.

What happens for $T$ just above $1$. With no equilibrium left, every orbit must go somewhere, and on the cylinder the only remaining possibility is a rotating periodic orbit — a running solution, the pendulum turning at a steady average rate. Poincaré–Bendixson does not prove it (the cylinder is not simply connected and the theorem's plane version does not apply), but the index count does the negative half: no contractible cycle can exist, since there is nothing of index $+1$ to enclose.

Hysteresis, for the curious. For small $\gamma$ the running solution persists as $T$ is lowered back below $1$, down to a second threshold $T_c(\gamma) < 1$, so the system remembers which way it came: two stable states coexist in between. This is the driven damped pendulum's celebrated bistability, and it is the mechanical analogue of the Josephson junction's current–voltage curve — the same equation, with $\theta$ the superconducting phase difference.

02 · bifurcations · example

The torque, drawn: where $\sin\theta = T$

Left: rest needs $\sin\theta = T$. At $T = 0.8$ the line cuts the sine twice, at $\theta_1 = 53.13°$ (the sink) and $\theta_2 = 126.87°$ (the saddle); at $T = 1$ it touches once at $\pi/2$; at $T = 1.2$ it misses. Right: the same, as $T$ moves — the sideways parabola of "Families, and where the portrait breaks", met in the wild.

checks

Limit $T \to 0$: $\theta_1 \to 0$, $\theta_2 \to \pi$ — the original two equilibria.
Nearing threshold: $\beta = \tfrac12\sqrt{4\cos\theta_1-\gamma^2}$ hits $0$ at $\cos\theta_1 = \gamma^2/4$, i.e. $T = 0.9980$ here: past that the sink is a node, so the colliding pair is real and merges along the $\theta$ axis.
Dimensions: $T$ is a torque over $mgl$, a pure number, so comparing it with $\sin\theta$ is legitimate.

What to look for. On the left the sink and the saddle are the two crossings of one horizontal line with one hump of the sine. Raising $T$ slides the line up and the crossings towards each other; they merge where the line is tangent, at the top of the hump, and that tangency is the non-hyperbolic moment: $\cos\theta = 0$ there, so $\det Df = 0$. On the right, near $T = 1$, $\theta_{1,2} = \pi/2 \mp \sqrt{2(1 - T)} + \dots$ (expand $\sin\theta = 1 - \tfrac12(\theta - \pi/2)^2 + \dots$), so the two branches meet like $\pm\sqrt{1-T}$ — exactly the saddle-node normal form with $\mu = 1 - T$.

The numbers. $\arcsin 0.8 = 0.9273$ rad $= 53.13°$ and $\pi - 0.9273 = 2.2143$ rad $= 126.87°$, as on the previous slide. The plots are computed from these formulas, not integrated.

02 · closing

One pendulum, six instruments

question	tool	what it gave, exactly
Where can it rest?	solve $f = 0$	two points on the cylinder: hanging and upright.
What happens to a nudge?	linearisation + Hartman–Grobman	$\gamma=\tfrac12$: stable spiral $-0.25\pm0.9682\,i$ below, saddle $0.7808 / -1.2808$ above.
And when $\gamma = 0$?	Lyapunov with the energy	a centre, stable but not asymptotically — the case linearisation cannot decide.
Where does it end up?	LaSalle on $\{V < 2\}$	every orbit inside the separatrix converges to the hanging position.
Can it repeat itself?	$\int_0^T\dot V = -\gamma\int\omega^2$	no periodic orbit at all for $\gamma>0$, swinging or rotating.
Which cycles are even possible?	the index	a contractible cycle must enclose the $+1$ point; rotations bound no disc and escape the count.
What if we drive it?	saddle-node at $T=1$	the $+1$ and the $-1$ collide and annihilate; beyond it, no rest at all.

Not one line of $\theta(t)$ was ever written down, because none exists. Everything above is exact.

The row that matters most is the fourth. A basin is a global statement about a nonlinear system with no closed-form solution, obtained from one guessed function and two compactness arguments. That is the technique to carry forward: in unit 14 the same move gives Liouville's theorem, in unit 17 it gives the invariant tori, and in unit 39 the Friedmann equations' qualitative behaviour is read off a potential in exactly this way.

What this unit deliberately did not do: compute anything approximately. There are no numerical trajectories in the argument (only in one figure, flagged as such), no perturbation series (unit 21), and no claims about how long convergence takes. The qualitative theory answers "what, and where to", never "when" — and knowing which questions a method cannot answer is half of knowing the method.

02 · checkpoint

Done when

checkpoint

Given a planar nonlinear system you have not seen before, you:

locate and classify every equilibrium — solve $f=0$, compute $Df$, read $\tau$ and $\Delta$;
decide at each whether linearisation is conclusive, by checking hyperbolicity, and say which theorem you are invoking;
construct a Lyapunov function where it is not, and state what it proves — stability, asymptotic stability, or a basin;
either prove or rule out a periodic orbit, using a trapping region with Poincaré–Bendixson, or the index theorem, or a strictly monotone quantity.

Failing it is unambiguous: an unclassified equilibrium, a linearisation trusted at a centre, or a cycle neither produced nor excluded.

Reading

Strogatz, Nonlinear Dynamics and Chaos, ch. 5–8 — the same material, slower · Perko, Differential Equations and Dynamical Systems, ch. 2–4 — the proofs cited here · Hirsch, Smale & Devaney, ch. 4–10 (optional).

06 the damped driven oscillator, which linearises everything on this deck and then solves it · 03 multivariable calculus, done properly · 14 where the conserved function becomes a symplectic form · 17 where Poincaré–Bendixson runs out.

./mr problems 02 — twelve problems, four hours. hints one rung at a time and the worked solution under each problem, after your attempt; ./mr done 02 when the checkpoint holds.

Try the checkpoint on this before the problem sheet: $\dot x = y$, $\dot y = -x + y(1 - x^2 - 2y^2)$. Equilibrium at the origin only; $Df(0,0) = \begin{pmatrix}0&1\\-1&1\end{pmatrix}$, $\tau = 1$, $\Delta = 1$, an unstable spiral, hyperbolic, so linearisation is conclusive there. Then look for a trapping annulus: $\frac{\mathrm{d}}{\mathrm{d}t}\tfrac12(x^2+y^2) = y^2(1-x^2-2y^2)$, which is $\le 0$ once $x^2 + 2y^2 \ge 1$ and $\ge 0$ once $x^2+2y^2 \le 1$ — so the annulus between the circles of radius $1/\sqrt2$ and $1$ traps, contains no equilibrium, and Poincaré–Bendixson gives a cycle. Problem P6 is the harder version of the same three steps.

If step 3 is the one that feels impossible, that is the correct feeling: there is no method, only a small catalogue of guesses — the energy, a quadratic form $\mathbf x^\top P\mathbf x$ solving a linear equation for $P$, a sum of squares. Problems P4 and P5 exercise the first two.

Phase portraits: what you know when there is no formula

Six questions, one pendulum

Vector fields, orbits and flows

No formula, and that is the normal case

An autonomous system is a vector field

The pendulum's field, drawn

Orbits cannot cross

The flow is a one-parameter group

The pendulum's state space is a cylinder

Two flows with extra structure

The undamped pendulum, drawn as level curves

Equilibria and linearisation

The derivative of a map, in one slide

Linearisation: what it is, and what it claims

Planar linear systems: two numbers decide

Solving $\dot{\boldsymbol\xi} = A\boldsymbol\xi$, all three cases

The cases, drawn on the pendulum's own matrices

The pendulum's two equilibria, classified

Hartman–Grobman, and the hole in it

Hartman–Grobman at the pendulum's saddle

When linearisation lies

Three systems, one linearisation

Lyapunov's direct method

Stability, defined

Differentiating along a solution

Lyapunov's theorem

Lyapunov's proof, as a picture

The undamped pendulum: a centre, proved stable

Where orbits end up: $\omega$-limit sets

LaSalle's invariance principle

The damped pendulum stops, and we can say where from

LaSalle's argument, drawn

Periodic orbits

Periodic orbits and limit cycles

Poincaré–Bendixson

Why the plane: crossings of a transverse segment

A trapping annulus, with the inequalities shown

Ruling cycles out

The index of a vector field

The index of a closed curve

Reading an index off the arrows

Indices of equilibria, and what they count

The pendulum: $+1$, $-1$, and the orbits that escape the theorem

Bifurcations

Families, and where the portrait breaks

Transcritical and pitchfork

Hopf: an equilibrium trades stability for a cycle

Hopf, drawn on both sides of the threshold

The pendulum with a constant torque

The torque, drawn: where $\sin\theta = T$

One pendulum, six instruments

Done when

Reading

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